RD Sharma XI 2020 2021 Volume 1 Solutions for Class 11 Commerce Maths Chapter 10 - Sine And Cosine Formulae And Their Applications

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RD Sharma XI 2020 2021 Volume 1 Solutions for Class 11 Commerce Maths Chapter 10 Sine And Cosine Formulae And Their Applications are provided here with simple step-by-step explanations. These solutions for Sine And Cosine Formulae And Their Applications are extremely popular among class 11 Commerce students for Maths Sine And Cosine Formulae And Their Applications Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma XI 2020 2021 Volume 1 Book of class 11 Commerce Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma XI 2020 2021 Volume 1 Solutions. All RD Sharma XI 2020 2021 Volume 1 Solutions for class 11 Commerce Maths are prepared by experts and are 100% accurate.

Page No 10.12:

Question 1:

If in âˆ†ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

$\frac{a}{\sin 45 °} = \frac{b}{\sin 60 °} = \frac{c}{\sin 75 °} = k \Rightarrow \frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{1}{2 \sqrt{2}} (1 + \sqrt{3})} [∵ \sin 75 ° = \sin (30 ° + 45 °) = \sin 30 ° \cos 45 ° + \sin 45 ° \cos 30 °]$

On multiplying by $2 \sqrt{2}$ , we get:
$a : b : c = 2 : \sqrt{6} : (1 + \sqrt{3})$

Hence, the ratio of the sides is $2 : \sqrt{6} : (1 + \sqrt{3})$ .

Page No 10.12:

Question 2:

If in âˆ†ABC, ∠C = 105°, ∠B = 45° and a = 2, then find b.

Answer:

$We know, A + B + C = π ∴ A = π - (B + C) \Rightarrow A = 180 ° - (45 ° + 105 °) = 30 ° Now, According to sine rule, \frac{a}{sinA} = \frac{b}{sinB} . \Rightarrow \frac{2}{\sin 30 °} = \frac{b}{\sin 45 °} (∵ a = 2, ∠ B = 45^{°}) \Rightarrow \frac{2}{\frac{1}{2}} = \frac{b}{\frac{1}{\sqrt{2}}} \Rightarrow 4 \times \frac{1}{\sqrt{2}} = b \Rightarrow b = 2 \sqrt{2}$

Page No 10.12:

Question 3:

In âˆ†ABC, if a = 18, b = 24 and c = 30 and ∠c = 90°, find sin A, sin B and sin C.

Answer:

Given,∠C = 90°, a = 18, b = 24 and c = 30

According to sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ .

$\Rightarrow \frac{c}{\sin C} = \frac{a}{\sin A} \Rightarrow \sin A = \frac{a \sin C}{c} = \frac{18 \times \sin (90 °)}{30} = \frac{18}{30} = \frac{3}{5} Also, \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow \sin B = \frac{b \sin C}{c} = \frac{24 \sin 90 °}{30} = \frac{24}{30} = \frac{4}{5} and \sin C = \sin 90 ° = 1$

Page No 10.12:

Question 4:

In triangle ABC, prove the following:

$\frac{a - b}{a + b} = \frac{\tan (\frac{A - B}{2})}{\tan (\frac{A + B}{2})}$

Answer:

$Assume \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Consider the LHS of the equation $\frac{a - b}{a + b} = \frac{\tan (\frac{A - B}{2})}{\tan (\frac{A + B}{2})}$ .
$LHS = \frac{a - b}{a + b} = \frac{k (\sin A - \sin B)}{k (\sin A + \sin B)}$

$∵ \sin A - \sin B = 2 \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}), \sin A + \sin B = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})$ .

$∴ LHS = \frac{2 \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2})}{2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})} = \frac{\tan (\frac{A - B}{2})}{\tan (\frac{A + B}{2})} = RHS Hence proved .$

Page No 10.13:

Question 5:

In triangle ABC, prove the following:

$(a - b) \cos \frac{C}{2} = c \sin (\frac{A - B}{2})$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ ...(1)

Consider the LHS of the equation $(a - b) \cos \frac{C}{2} = c \sin (\frac{A - B}{2})$
$LHS = (a - b) \cos \frac{C}{2} = k (\sin A - \sin B) \cos \frac{C}{2} (using (1)) = k \times 2 \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}) \cos \frac{C}{2} = 2 k \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}) \cos (\frac{π - (A + B)}{2}) [∵ A + B + C = π] = 2 k \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2}) \sin (\frac{A + B}{2}) = k \sin (\frac{A - B}{2}) \sin (A + B) [∵ 2 \cos (\frac{A + B}{2}) \sin (\frac{A + B}{2}) = \sin (A + B)] = k \sin \frac{A - B}{2} \sin (π - C) [∵ A + B + C = π] = k \sin C \sin (\frac{A - B}{2}) = C \sin (\frac{A - B}{2}) = RHS$

Hence proved.

Page No 10.13:

Question 6:

In triangle ABC, prove the following:

$\frac{c}{a - b} = \frac{\tan (\frac{A}{2}) + \tan (\frac{B}{2})}{\tan (\frac{A}{2}) - \tan (\frac{B}{2})}$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ ...(1)

We need to prove:
$\frac{c}{a - b} = \frac{\tan (\frac{A}{2}) + \tan (\frac{B}{2})}{\tan (\frac{A}{2}) - \tan (\frac{B}{2})}$

Consider

$LHS = \frac{c}{a - b} = \frac{k \sin C}{k (\sin A - \sin B)} (using (1)) = \frac{2 \sin \frac{C}{2} \cos \frac{C}{2}}{2 \sin (\frac{A - B}{2}) \cos (\frac{A + B}{2})} = \frac{\sin (\frac{π - (A + B)}{2}) \cos \frac{C}{2}}{\sin (\frac{A - B}{2}) \cos (\frac{A + B}{2})} (∵ A + B + C = π) = \frac{\cos \frac{C}{2} \cos (\frac{A + B}{2})}{\sin (\frac{A - B}{2}) \cos (\frac{A + B}{2})} = \frac{\cos \frac{C}{2}}{\sin (\frac{A - B}{2})} . . . (2)$

$RHS = \frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{\tan \frac{A}{2} - \tan \frac{B}{2}} = \frac{\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} + \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}}{\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} - \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}} = \frac{\sin \frac{A}{2} \cos \frac{B}{2} + \sin \frac{B}{2} \cos \frac{A}{2}}{\sin \frac{A}{2} \cos \frac{B}{2} - \sin \frac{B}{2} \cos \frac{A}{2}} = \frac{\sin (\frac{A + B}{2})}{\sin (\frac{A - B}{2})} = \frac{\sin (\frac{π - C}{2})}{\sin (\frac{A - B}{2})} = \frac{\cos \frac{C}{2}}{\sin (\frac{A - B}{2})} = LHS (from (2)) Hence proved .$

Page No 10.13:

Question 7:

In triangle ABC, prove the following:

$\frac{c}{a + b} = \frac{1 - \tan (\frac{A}{2}) \tan (\frac{B}{2})}{1 + \tan (\frac{A}{2}) \tan (\frac{B}{2})}$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ ...(1)

We need to prove:
$\frac{c}{a + b} = \frac{1 - \tan (\frac{A}{2}) \tan (\frac{B}{2})}{1 + \tan (\frac{A}{2}) \tan (\frac{B}{2})}$

Consider

$LHS = \frac{c}{a + b} = \frac{ksinC}{k (sinA + sinB)} (using (1)) = \frac{2 \sin \frac{C}{2} \cos \frac{C}{2}}{2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})} = \frac{\sin \frac{C}{2} \cos (\frac{π - (A + B)}{2})}{\sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})} (∵ A + B + C = π) = \frac{\sin \frac{C}{2} \sin (\frac{A + B}{2})}{\sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})} = \frac{\sin \frac{C}{2}}{\cos (\frac{A - B}{2})} . . . (2) RHS = \frac{1 - t a n \frac{A}{2} t a n \frac{B}{2}}{1 + t a n \frac{A}{2} t a n \frac{B}{2}} = \frac{1 - \frac{s i n \frac{A}{2}}{c o s \frac{A}{2}} \frac{s i n \frac{B}{2}}{c o s \frac{B}{2}}}{1 + \frac{s i n \frac{A}{2}}{c o s \frac{A}{2}} \frac{s i n \frac{B}{2}}{c o s \frac{B}{2}}} = \frac{c o s \frac{A}{2} c o s \frac{B}{2} - s i n \frac{A}{2} s i n \frac{B}{2}}{c o s \frac{A}{2} c o s \frac{B}{2} + s i n \frac{A}{2} s i n \frac{B}{2}} = \frac{c o s (\frac{A + B}{2})}{c o s (\frac{A - B}{2})} = \frac{c o s (\frac{π - C}{2})}{c o s (\frac{A - B}{2})} (∵ A + B + C = π) = \frac{\sin \frac{C}{2}}{\cos (\frac{A - B}{2})} = LHS (from (2)) Hence proved .$

Page No 10.13:

Question 8:

In triangle ABC, prove the following:

$\frac{a + b}{c} = \frac{\cos (\frac{A - B}{2})}{\sin \frac{C}{2}}$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ ...(1)

Then,
Consider the LHS of the equation $\frac{a + b}{c} = \frac{\cos (\frac{A - B}{2})}{\sin \frac{C}{2}}$ .
$LHS = \frac{a + b}{c} = \frac{k \sin A + k \sin B}{k \sin C} (using (1)) = \frac{2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})}{2 \sin \frac{C}{2} \cos \frac{C}{2}} = \frac{\sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})}{\sin \frac{C}{2} \cos (\frac{π - (A + B)}{2})} (∵ A + B + C = π) = \frac{\sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})}{\sin \frac{C}{2} \sin (\frac{A + B}{2})} = \frac{\cos (\frac{A - B}{2})}{\sin \frac{C}{2}} = RHS Hence proved .$

Page No 10.13:

Question 9:

In any triangle ABC, prove the following:

$\sin (\frac{B - C}{2}) = \frac{b - c}{a} \cos \frac{A}{2}$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,
Consider the RHS of the equation $\sin (\frac{B - C}{2}) = \frac{b - c}{a} \cos \frac{A}{2}$
$RHS = \frac{b - c}{a} \cos \frac{A}{2} = \frac{k (\sin B - \sin C)}{k \sin A} \cos (\frac{π - (B + C)}{2}) (∵ A + B + C = π) = \frac{2 \sin (\frac{B - C}{2}) \cos (\frac{B + C}{2})}{\sin A} = \frac{\sin (\frac{B - C}{2}) 2 \cos (\frac{B + C}{2})}{\sin A} \sin (\frac{B + C}{2}) = \frac{\sin (\frac{B - C}{2}) \sin (B + C)}{\sin A} = \frac{\sin (\frac{B - C}{2}) \sin (π - A)}{\sin A} = \frac{\sin A \sin (\frac{B - C}{2})}{\sin A} = \sin (\frac{B - C}{2}) = LHS Hence proved .$

Page No 10.13:

Question 10:

In triangle ABC, prove the following:

$\frac{a^{2} - c^{2}}{b^{2}} = \frac{\sin (A - C)}{\sin (A + C)}$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$
Then,
Consider the LHS of the equation $\frac{a^{2} - c^{2}}{b^{2}} = \frac{\sin (A - C)}{\sin (A + C)}$ .
$LHS = \frac{{[k \sin (A)]}^{2} - {[k \sin (C)]}^{2}}{{(k \sin (B))}^{2}} = \frac{k^{2} (\sin^{2} (A) - \sin^{2} (C))}{k^{2} \sin^{2} (B)} = \frac{\sin (A + C) \sin (A - C)}{\sin^{2} B} [∵ \sin^{2} A - \sin^{2} C = \sin (A + C) \sin (A - C)] = \frac{\sin (A + C) \sin (A - C)}{S i n^{2} (π - (A + C))} [∵ (A + B + C) = π] = \frac{\sin (A + C) \sin (A - C)}{\sin^{2} (A + C)} = \frac{\sin (A - C)}{\sin (A + C)} = RHS Hence proved .$

Page No 10.13:

Question 11:

In triangle ABC, prove the following:
$b \sin B - c \sin C = a \sin (B - C)$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$
Then,

Consider the LHS of he equation $b \sin B - c \sin C = a \sin (B - C)$ .
LHS $= k \sin B \sin B - k \sin C \sin C$
$= k (\sin^{2} B - \sin^{2} C) = k [\sin (B + C) \sin (B - C)] [∵ \sin^{2} B - \sin^{2} C = \sin (B + C) \sin (B - C)] = k [\sin (π - A) \sin (B - C)] [∵ A + B + C = π] = k \sin A \sin (B - C) [∵ a = k \sin A] = a \sin (B - C) = RHS Hence proved .$

Page No 10.13:

Question 12:

In triangle ABC, prove the following:
$a^{2} \sin (B - C) = (b^{2} - c^{2}) \sin A$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$
Then,

Consider the RHS of the equation $a^{2} \sin (B - C) = (b^{2} - c^{2}) \sin A$ .

$RHS = k^{2} \sin A (\sin^{2} B - \sin^{2} C) = k^{2} \sin A [\sin (B + C) \sin (B - C)] [∵ \sin^{2} B - \sin^{2} C = \sin (B + C) \sin (B - C)] = k^{2} \sin A [\sin (π - A) \sin (B - C)] [∵ A + B + C = π] = k^{2} \sin A [\sin (A) \sin (B - C)] = k^{2} \sin^{2} A \sin (B - C) = a^{2} \sin (B - C) = LHS [∵ a = k \sin A] Hence proved .$

Page No 10.13:

Question 13:

In triangle ABC, prove the following:

$\frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{\sin B}} = \frac{a + b - 2 \sqrt{a b}}{a - b}$

Answer:

Consider the LHS of the equation $\frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{\sin B}} = \frac{a + b - 2 \sqrt{a b}}{a - b}$ .

$LHS = \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{s i n B}} = \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{s i n B}} \times \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} - \sqrt{\sin B}} = \frac{\sin A + \sin B - (2 \times \sqrt{\sin A \sin B})}{\sin A - \sin B}$

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,
$LHS = \frac{\frac{a}{k} + \frac{b}{k} - 2 \times \sqrt{\frac{a}{k} \frac{b}{k}}}{\frac{a}{k} - \frac{b}{k}} = \frac{\frac{1}{k} (a + b - 2 \times \sqrt{a b})}{\frac{1}{k} (a - b)} = \frac{a + b - 2 \sqrt{a b}}{a - b} = R H S H e n c e p r o v e d .$

Page No 10.13:

Question 14:

In triangle ABC, prove the following:
$a (\sin B - \sin C) + (\sin C - \sin A) + c (\sin A - \sin B) = 0$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

Consider the LHS of the equation $a (\sin B - \sin C) + (\sin C - \sin A) + c (\sin A - \sin B) = 0$ .

$LHS = a (\sin B - \sin C) + b (\sin C - \sin A) + c (s i n A - s i n B) = k \sin A (\sin B - \sin C) + k \sin B (\sin C - \sin A) + k \sin C (s i n A - s i n B) = k \sin A \sin B - k \sin A \sin C + k \sin B \sin C - k \sin B \sin A + k s i n C \sin A - k \sin C \sin B = 0 = RHS Hence proved .$

Page No 10.13:

Question 15:

In triangle ABC, prove the following:

$\frac{a^{2} \sin (B - C)}{\sin A} + \frac{b^{2} \sin (C - A)}{\sin B} + \frac{c^{2} \sin (A - B)}{\sin C} = 0$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

Consider the LHS of the equation $\frac{a^{2} \sin (B - C)}{\sin A} + \frac{b^{2} \sin (C - A)}{\sin B} + \frac{c^{2} \sin (A - B)}{\sin C} = 0$ .

$LHS = \frac{a^{2} \sin (B - C)}{\sin A} + \frac{b^{2} \sin (C - A)}{\sin B} + \frac{c^{2} \sin (A - B)}{\sin C} = \frac{k^{2} \sin^{2} A \sin (B - C)}{\sin A} + \frac{k^{2} \sin^{2} B \sin (C - A)}{\sin B} + \frac{k^{2} \sin^{2} C \sin (A - B)}{\sin C} = k^{2} \sin A \sin (B - C) + k^{2} \sin B \sin (C - A) + k^{2} \sin C \sin (A - B) = k^{2} [\sin A (\sin B \cos C - \sin C \cos B) + \sin B (\sin C \cos A - \sin A \cos C) + \sin C (\sin A \cos B - \sin B \cos A)] = k^{2} (\sin A \sin B \cos C - \sin A \sin C \cos B + \sin B \sin C \cos A - \sin A \sin B \cos C + \sin A \sin C \cos B - \sin C \sin B \cos A) = 0 = RHS Hence proved .$

Page No 10.13:

Question 16:

In triangle ABC, prove the following:
$a^{2} (\cos^{2} B - \cos^{2} C) + b^{2} (\cos^{2} C - \cos^{2} A) + c^{2} (\cos^{2} A - \cos^{2} B) = 0$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

Consider the LHS of the equation $a^{2} (\cos^{2} B - \cos^{2} C) + b^{2} (\cos^{2} C - \cos^{2} A) + c^{2} (\cos^{2} A - \cos^{2} B) = 0$ .

$LHS = a^{2} (\cos^{2} B - \cos^{2} C) + b^{2} (\cos^{2} C - \cos^{2} A) + c^{2} (\cos^{2} A - \cos^{2} B) = k^{2} \sin^{2} A (1 - \sin^{2} B - 1 + \sin^{2} C) + k^{2} \sin^{2} B (1 - \sin^{2} C - 1 + \sin^{2} A) + k^{2} \sin^{2} C (1 - \sin^{2} A - 1 + \sin^{2} B) = k^{2} \sin^{2} A (\sin^{2} C - \sin^{2} B) + k^{2} \sin^{2} B (\sin^{2} A - \sin^{2} C) + k^{2} \sin^{2} C (\sin^{2} B - \sin^{2} A) = k^{2} (\sin^{2} A \sin^{2} C - \sin^{2} A \sin^{2} B + \sin^{2} A \sin^{2} B - \sin^{2} B \sin^{2} C + \sin^{2} C \sin^{2} B - \sin^{2} C \sin^{2} A) = k^{2} \times 0 = 0 = RHS Hence proved .$

Page No 10.13:

Question 17:

In triangle ABC, prove the following:
$b \cos B + c \cos C = a \cos (B - C)$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

Consider the LHS of the equation $b \cos B + c \cos C = a \cos (B - C)$ .

$LHS = b \cos B + c \cos C = k (\sin B \cos B + \sin C \cos C) = \frac{k}{2} (2 sinBcos B + 2 sinCcos C) = \frac{k}{2} (\sin 2 B + \sin 2 C) . . . (1) RHS = a \cos (B - C) = k \sin A \cos (B - C) = \frac{k}{2} [2 \sin A \cos (B - C)] = \frac{k}{2} [\sin (A + B - C) + \sin (A - B + C)] [∵ 2 \sin A \cos B = \sin (A + B) + \sin (A - B)] = \frac{k}{2} [\sin (π - C - C) + \sin (π - B - B)] [∵ \sin (π - A) = \sin A, A + B + C = π] = \frac{k}{2} (\sin 2 C + \sin 2 B) = \frac{k}{2} (\sin 2 B + \sin 2 C) = LHS [from (1)] Hence proved .$

Page No 10.13:

Question 18:

In triangle ABC, prove the following:

$\frac{\cos 2 A}{a^{2}} - \frac{\cos 2 B}{b^{2}} - \frac{1}{a^{2}} - \frac{1}{b^{2}}$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

Consider the LHS of the equation $\frac{\cos 2 A}{a^{2}} - \frac{\cos 2 B}{b^{2}} - \frac{1}{a^{2}} - \frac{1}{b^{2}}$ .

$LHS = \frac{\cos 2 A}{a^{2}} - \frac{\cos 2 B}{b^{2}} = \frac{1 - 2 \sin^{2} A}{a^{2}} - \frac{1 - 2 \sin^{2} B}{b^{2}} = \frac{1 - 2 \frac{a^{2}}{k^{2}}}{a^{2}} - \frac{1 - 2 \frac{b^{2}}{k^{2}}}{b^{2}} = \frac{\frac{k^{2} - 2 a^{2}}{k^{2}}}{a^{2}} - \frac{\frac{k^{2} - 2 b^{2}}{k^{2}}}{b^{2}} = \frac{k^{2} b^{2} - 2 a^{2} b^{2} - k^{2} a^{2} + 2 a^{2} b^{2}}{a^{2} b^{2}} = \frac{k^{2} (b^{2} - a^{2})}{k^{2} a^{2} b^{2}} = \frac{1}{a^{2}} - \frac{1}{b^{2}} = RHS Hence proved .$

Page No 10.13:

Question 19:

In triangle ABC, prove the following:

$\frac{\cos^{2} B - \cos^{2} C}{b + c} + \frac{\cos^{2} C - \cos^{2} A}{c + a} + \frac{c o s^{2} A - \cos^{2} B}{a + b} = 0$

Answer:

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

Consider the LHS of the equation $\frac{\cos^{2} B - \cos^{2} C}{b + c} + \frac{\cos^{2} C - \cos^{2} A}{c + a} + \frac{c o s^{2} A - \cos^{2} B}{a + b} = 0$ .

$LHS = \frac{\cos^{2} B - \cos^{2} C}{b + c} + \frac{\cos^{2} C - \cos^{2} A}{c + a} + \frac{\cos^{2} A - \cos^{2} B}{a + b} Now, \frac{\cos^{2} B - \cos^{2} C}{b + c} = \frac{\cos^{2} B - \cos^{2} C}{k (\sin B + \sin C)} = \frac{(\cos B + \cos C) (\cos B - c o s C)}{k (\sin B + \sin C)} (∵ \cos^{2} B - \cos^{2} C = (\cos B + \cos C) (\cos B - \cos C)) = \frac{[2 \cos (\frac{B + C}{2}) \cos (\frac{B - C}{2})] [- 2 \sin (\frac{B + C}{2}) \sin (\frac{B - C}{2})]}{2 k \sin (\frac{B + C}{2}) \sin (\frac{B - C}{2})} = \frac{- 2 \cos (\frac{B + C}{2}) \sin (\frac{B - C}{2})}{k} = \frac{- (\sin B - \sin C)}{k} = \frac{\sin C - \sin B}{k}$
Also,

$\frac{\cos^{2} C - \cos^{2} A}{c + a} = \frac{\cos^{2} C - \cos^{2} A}{k (\sin C + \sin A)} = \frac{(\cos C + \cos A) (\cos C - \cos A)}{k (\sin C + \sin A)} = \frac{[2 \cos (\frac{C + A}{2}) \cos (\frac{C - A}{2})] [- 2 \sin (\frac{C + A}{2}) \sin (\frac{C - A}{2})]}{2 k (\sin C + \sin A) k} = \frac{- 2 \cos (\frac{C + A}{2}) \cos (\frac{C - A}{2})}{k} = \frac{- (\sin C - \sin A)}{k} = \frac{\sin A - \sin C}{k}$
Similarly,
$\frac{\cos^{2} A - \cos^{2} B}{a + b} = \frac{\sin B - \sin A}{k}$

Thus,

$LHS = \frac{\sin A - \sin C}{k} + \frac{\sin B - \sin A}{k} + \frac{\sin C - \sin B}{k} = 0 = RHS$

Hence, in any triangle ABC, $\frac{\cos^{2} B - \cos^{2} C}{b + c} + \frac{\cos^{2} C - \cos^{2} A}{c + a} + \frac{c o s^{2} A - \cos^{2} B}{a + b} = 0$ .

Page No 10.13:

Question 20:

In âˆ†ABC, prove that:

$a \sin \frac{A}{2} \sin (\frac{B - C}{2}) + b \sin \frac{B}{2} \sin (\frac{C - A}{2}) + c \sin \frac{C}{2} \sin (\frac{A - B}{2}) = 0$ .

Answer:

Consider
$a \sin \frac{A}{2} \sin (\frac{B - C}{2}) + b \sin \frac{B}{2} \sin (\frac{C - A}{2}) + c \sin \frac{C}{2} \sin (\frac{A - B}{2})$
$= k [\sin A \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin B \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin C \sin \frac{C}{2} \sin (\frac{A - B}{2})] = k [\sin \{π - (B + C)\} \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin \{π - (C + A)\} \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin \{π - (A + B)\} \sin \frac{C}{2} \sin (\frac{A - B}{2})] (∵ A + B + C = π) = k [\sin (B + C) \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin (A + C) \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin (A + B) \sin \frac{C}{2} \sin (\frac{A - B}{2})] = k [2 \sin (\frac{B + C}{2}) \cos (\frac{B - C}{2}) \sin \frac{A}{2} \sin (\frac{B - C}{2}) + 2 \sin (\frac{A + C}{2}) \cos (\frac{C - A}{2}) \sin \frac{B}{2} \sin (\frac{C - A}{2}) + 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2}) \sin \frac{C}{2} \sin (\frac{A - B}{2})] = 2 k [\sin (\frac{B + C}{2}) \sin \frac{A}{2} \sin \frac{A}{2} \sin (\frac{B - C}{2}) + \sin (\frac{A + C}{2}) \sin \frac{B}{2} \sin \frac{B}{2} \sin (\frac{C - A}{2}) + \sin (\frac{A + B}{2}) \sin \frac{C}{2} \sin \frac{C}{2} \sin (\frac{A - B}{2})] = 2 k [\sin (\frac{B + C}{2}) \sin (\frac{B - C}{2}) \sin^{2} \frac{A}{2} + \sin (\frac{A + C}{2}) \sin (\frac{C - A}{2}) \sin^{2} \frac{B}{2} + \sin (\frac{A + B}{2}) \sin (\frac{A - B}{2}) \sin^{2} \frac{C}{2}] = 2 k \sin^{2} \frac{A}{2} (\sin^{2} \frac{B}{2} - \sin^{2} \frac{C}{2}) + 2 k \sin^{2} \frac{B}{2} (\sin^{2} \frac{C}{2} - \sin^{2} \frac{A}{2}) + 2 k \sin^{2} \frac{C}{2} (\sin^{2} \frac{A}{2} - \sin^{2} \frac{B}{2}) = 2 k (\sin^{2} \frac{A}{2} \sin^{2} \frac{B}{2} - \sin^{2} \frac{A}{2} \sin^{2} \frac{C}{2} + \sin^{2} \frac{B}{2} \sin^{2} \frac{C}{2} - \sin^{2} \frac{A}{2} \sin^{2} \frac{B}{2} + \sin^{2} \frac{A}{2} \sin^{2} \frac{C}{2} - \sin^{2} \frac{C}{2} \sin^{2} \frac{B}{2}) = k (0) = 0$

Hence proved.

Page No 10.13:

Question 21:

In âˆ†ABC, prove that:

$\frac{b \sec B + c \sec C}{\tan B + \tan C} = \frac{c \sec C + a \sec A}{\tan C + \tan A} = \frac{a \sec A + b \sec B}{\tan A + \tan B}$

Answer:

Let ABC be any triangle.

Suppose $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Now,
$\frac{b \sec B + c \sec C}{\tan B + \tan C} = \frac{\frac{b}{\cos B} + \frac{c}{\cos C}}{\frac{\sin B}{\cos B} + \frac{\sin C}{\cos C}} = \frac{b \cos C + c \cos B}{\sin B \cos C + \sin C \cos B} = \frac{k \sin B \cos C + k \sin C \cos B}{\sin B \cos C + \sin C \cos B} = \frac{k (\sin B \cos C + \sin C \cos B)}{\sin B \cos C + \sin C \cos B} = k . . . (1)$
Also,
$\frac{c \sec C + a \sec A}{\tan C + \tan A} = \frac{\frac{c}{\cos C} + \frac{a}{\cos A}}{\frac{\sin C}{\cos C} + \frac{\sin A}{\cos A}} = \frac{c \cos A + a \cos C}{\sin C \cos A + \sin A \cos C} = \frac{k \sin C \cos A + k \sin A \cos C}{\sin C \cos A + \sin A \cos C} = \frac{k (\sin C \cos A + \sin A \cos C)}{\sin C \cos A + \sin A \cos C} = k . . . (2) and \frac{a \sec A + b \sec B}{\tan A + \tan B} = \frac{\frac{a}{\cos A} + \frac{b}{\cos B}}{\frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}} = \frac{a \cos B + b \cos A}{\sin A \cos B + \sin B \cos A} = \frac{k (\sin A \cos B + \sin B \cos A)}{\sin A \cos B + \sin B \cos A} = k . . . (3)$

From (1), (2) and (3), we get:

$\frac{b \sec B + c \sec C}{\tan B + \tan C} = \frac{c \sec C + a \sec A}{\tan C + \tan A} = \frac{a \sec A + b \sec B}{\tan A + \tan B}$

Hence proved.

Page No 10.13:

Question 22:

In triangle ABC, prove the following:
$a \cos A + b \cos B + c \cos C = 2 b \sin A \sin C = 2 c \sin A \sin B$

Answer:

$Let \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k . . . (1) Consider the LHS of the equation a \cos A + b \cos B + c \cos C . a \cos A + b \cos B + c \cos C = k (\sin A \cos A + \sin B \cos B + \sin C \cos C) = \frac{k}{2} (2 s i n A c o s A + 2 s i n A c o s A + 2 s i n C c o s C) = \frac{k}{2} (\sin 2 A + \sin 2 B + \sin 2 C) = \frac{k}{2} [2 \sin (A + B) \cos (A - B) + 2 \sin C \cos C] = \frac{k}{2} [2 \sin (π - C) \cos (A - B) + 2 \sin C \cos C] = \frac{k}{2} [2 \sin C \cos (A - B) + 2 \sin C \cos C] = \frac{2 k \sin C}{2} [\cos (A - B) + \cos C]$
$= k \sin C [\cos (A - B) + \cos \{π - (A + B)\}] = k \sin C [\cos (A - B) - \cos (A + B)] = k \sin C [2 \sin A sinB] = 2 k \sin A \sin B \sin C . . . (1)$

$Now, on putting k \sin C = C in equation (1), we get : 2 c \sin A \sin B and on putting k \sin B = b in equation (1), we get : 2 b \sin A \sin C$

So, from (1), we have
$a \cos A + b \cos B + c \cos C = 2 b \sin A \sin C = 2 c \sin A \sin B$ .

Hence proved.

Page No 10.13:

Question 23:

$a (\cos B \cos C + \cos A) = b (\cos C \cos A + \cos B) = c (\cos A \cos B + \cos C)$

Answer:

Suppose $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Consider:
$a (\cos B \cos C + \cos A) = k \sin A (\cos B \cos C + \cos A) = k (\sin A \cos B \cos C + \cos A \sin A) = k [\frac{1}{2} \cos C \{\sin (A + B) + \sin (A - B)\} + \sin A \cos A] = k [\frac{1}{2} \{\sin (A + B) \cos C + \sin (A - B) \cos C\} + \sin A \cos A] = k [\frac{1}{2} \{\frac{1}{2} [\sin (A + B + C) + \sin (A + B - C) + \sin (A - B + C) + \sin (A - B - C)]\} + \sin A \cos A] = k [\frac{1}{4} \{sinπ + \sin (π - 2 C) + \sin (π - 2 B) - \sin (π - 2 A)\} + \frac{\sin 2 A}{2}] (∵ A + B + C = π) = \frac{k}{4} (\sin 2 C + \sin 2 B + \sin 2 A) . . . . (1) and b (\cos A \cos C + \cos B) = k (\sin B \cos A \cos C + sinBcos B) = k [\frac{1}{2} \cos A \{\sin (B + C) + \sin (B - C)\} + \frac{\sin 2 B}{2}] = k (\frac{1}{2} (\sin (B + C) \cos A + \sin (B - C) \cos A) + \frac{\sin 2 B}{2}) = k (\frac{1}{4} (\sin (B + C + A) + \sin (B + C - A) + \sin (B - C + A) + \sin (B - C - A)) + \frac{\sin 2 B}{2}) = \frac{k}{4} (sinπ + \sin (π - 2 A) + \sin (π - 2 C) - \sin (π - 2 B) + \frac{\sin 2 B}{2}) (∵ A + B + C = π) = \frac{k}{4} (\sin 2 A + \sin 2 C + \sin 2 B) . . . (2) Similarly, c (\cos A \cos B + \cos C) = \frac{k}{4} (\sin 2 A + \sin 2 B + \sin 2 C) . . . (3)$

From (1), (2) and (3), we get:

$a (\cos B \cos C + \cos A) = b (\cos C \cos A + \cos B) = c (\cos A \cos B + \cos C)$

Hence proved.

Page No 10.13:

Question 24:

In âˆ†ABC, prove that $a (\cos C - \cos B) = 2 (b - c) \cos^{2} \frac{A}{2} .$

Answer:

$Consider a (\cos C - \cos B) = k (\sin A \cos C - \sin A \cos B) [∵ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k] = \frac{k}{2} (2 \sin A \cos C - 2 \sin A \cos B) = \frac{k}{2} [\sin (A + C) + \sin (A - C) - \sin (A + B) - \sin (A - B)] = \frac{k}{2} [\sin (π - B) + \sin (A - C) - \sin (π - C) - \sin (A - B)] (∵ A + B + C = π) = \frac{k}{2} [\sin B - \sin C + \sin (A - C) - \sin (A - B)] = \frac{k}{2} [2 \sin (\frac{B - C}{2}) \cos (\frac{B + C}{2}) + 2 \sin (\frac{A - C - A + B}{2}) \cos (\frac{A - C + A - B}{2})] = k \sin (\frac{B - C}{2}) [\cos (\frac{π}{2} - \frac{A}{2}) + \cos \{\frac{2 A - (π - A)}{2}\}] = k \sin (\frac{B - C}{2}) (\sin \frac{A}{2} + \sin \frac{3 A}{2}) = k \sin (\frac{B - C}{2}) [2 \sin (\frac{\frac{A}{2} + \frac{3 A}{2}}{2}) \cos (\frac{\frac{3 A}{2} - \frac{A}{2}}{2})] = 2 k \sin (\frac{B - C}{2}) \sin A \cos \frac{A}{2} = 4 k \sin (\frac{B - C}{2}) \sin \frac{A}{2} \cos^{2} \frac{A}{2} . . . (1) Now, Consider 2 (b - c) \cos^{2} \frac{A}{2} = 2 k (\sin B - \sin C) \cos^{2} \frac{A}{2} = 2 k [2 \sin (\frac{B - C}{2}) \cos (\frac{B + C}{2})] \cos^{2} \frac{A}{2} = 4 k \sin (\frac{B - C}{2}) \cos (\frac{π}{2} - \frac{A}{2}) \cos^{2} \frac{A}{2} = 4 k \sin (\frac{B - C}{2}) \sin \frac{A}{2} \cos^{2} \frac{A}{2} . . . (2) From (1) & (2), we get a (\cos C - \cos B) = 2 (b - c) \cos^{2} \frac{A}{2} Hence proved .$

Page No 10.13:

Question 25:

In âˆ†ABC, prove that if θ be any angle, then b cosθ = c cos (A − θ) + a cos (C + θ).

Answer:

Suppose $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ . ...(1)

Consider the RHS of the equation b cosθ = c cos (A − θ) + a cos (C + θ).

$RHS = c \cos (A - θ) + a \cos (C + θ) = k \sin C \cos (A - θ) + k \sin A \cos (C + θ) (from (1)) = \frac{k}{2} [2 \sin C \cos (A - θ) + 2 \sin A \cos (C + θ)] = \frac{k}{2} [\sin (A + C - θ) + \sin (C + θ - A) + \sin (A + C + θ) + \sin (A - C - θ)] = \frac{k}{2} (\sin (π - B - θ) + \sin (C + θ - A) + \sin (π - B + θ) - \sin (C + θ - A)) (∵ A + B + C = π) = \frac{k}{2} (\sin (B + θ) + \sin (B - θ)) = \frac{k}{2} (\sin B \cos θ + \sin θ \cos B + \sin B \cos θ - \sin θ \cos B) = \frac{k}{2} (2 \sin B \cos θ) = k \sin B \cos θ = b \cos θ = LHS Hence proved .$

Page No 10.13:

Question 26:

In âˆ†ABC, if sin²A + sin²B = sin²C. show that the triangle is right-angled.

Answer:

In âˆ† ABC,
Given, $\sin^{2} A + \sin^{2} B = \sin^{2} C . . . . . . (1)$

Suppose $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ .

⇒ $\sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k}$

On putting these values in equation (1), we get:

$\frac{a^{2}}{k^{2}} + \frac{b^{2}}{k^{2}} = \frac{c^{2}}{k^{2}} \Rightarrow a^{2} + b^{2} = c^{2}$

Thus, âˆ† ABC is right-angled.

Page No 10.14:

Question 27:

In âˆ†ABC, if a², b² and c² are in A.P., prove that cot A, cot B and cot C are also in A.P.

Answer:

$Let \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$

Then,
$\sin A = k a, \sin B = k b, \sin C = k c$
a², b² and c² are in A.P.
$\Rightarrow 2 b^{2} = a^{2} + c^{2} \Rightarrow 2 (a^{2} + c^{2} - b^{2}) = 2 (2 b^{2} - b^{2}) = 2 b^{2} = b^{2} + b^{2} + c^{2} - a^{2} - c^{2} + a^{2} \Rightarrow 2 (a^{2} + c^{2} - b^{2}) = b^{2} + c^{2} - a^{2} + a^{2} + b^{2} - c^{2} \Rightarrow \frac{2 (a^{2} + c^{2} - b^{2})}{2 a b c} = \frac{(b^{2} + c^{2} - a^{2})}{2 a b c} + \frac{(a^{2} + b^{2} - c^{2})}{2 a b c} \Rightarrow \frac{2 \cos B}{k b} = \frac{\cos A}{k a} + \frac{\cos C}{k c} \Rightarrow \frac{2 \cos B}{\sin B} = \frac{\cos A}{\sin A} + \frac{\cos C}{\sin C} \Rightarrow 2 \cot B = \cot A + \cot C \Rightarrow \cot A, \cot B and \cot C are in AP .$

Page No 10.14:

Question 28:

The upper part of a tree broken by the wind makes an angle of 30° with the ground and the distance from the root to the point where the top of the tree touches the ground is 15 m. Using sine rule, find the height of the tree.

Answer:

Suppose BD be the tree and the upper part of the tree is broken over by the wind at point A.

$The total height of the tree is x + y . I n △ A B C, ∠ C = 30 ° a n d ∠ B = 90 ° . ∴ ∠ A = 60 ° . So, on using sine rule, we get: \frac{A B}{\sin 30 °} = \frac{B C}{\sin 60 °} = \frac{A C}{\sin 90 °} \Rightarrow \frac{x}{\sin 30 °} = \frac{15}{\sin 60 °} = \frac{y}{\sin 90 °} So, \frac{x}{\sin 30 °} = \frac{15}{\sin 60 °} \Rightarrow \frac{x}{\frac{1}{2}} = \frac{15}{\frac{\sqrt{3}}{2}} \Rightarrow x = \frac{15}{\sqrt{3}} = 5 \sqrt{3} Also, \frac{15}{\sin 60 °} = \frac{y}{\sin 90 °} \Rightarrow \frac{15}{\frac{\sqrt{3}}{2}} = y \Rightarrow y = \frac{30}{\sqrt{3}} = 10 \sqrt{3} So, the height of the tree is x + y = 5 \sqrt{3} + 10 \sqrt{3} m = 15 \sqrt{3} m$

Page No 10.14:

Question 29:

At the foot of a mountain, the elevation of it summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.

Answer:

Suppose, AB is a mountain of height t + x.

$In △ D F C, \sin 30 ° = \frac{x}{1000} \Rightarrow x = 1000 \times (\frac{1}{2}) = 500 m and \tan 30 ° = \frac{x}{y} \Rightarrow y = \frac{x}{\tan 30 °} = 500 \sqrt{3} In ∆ A B C, \tan 45 ° = \frac{t + x}{y + z} \Rightarrow t + x = y + z . . . (1) In ∆ A D E, \tan 60 ° = \frac{t}{z} \Rightarrow t = \sqrt{3} z . . . (2) From (1) and (2), we have \sqrt{3} z + x = y + z \Rightarrow z (\sqrt{3} - 1) = 500 (\sqrt{3} - 1) \Rightarrow z = 500 m ∴ t = \sqrt{3} z = 500 \sqrt{3}$

Hence, height of the mountain = $t + x = 500 \sqrt{3} + 500 = 500 (\sqrt{3} + 1) m$ .

Page No 10.14:

Question 30:

A person observes the angle of elevation of the peak of a hill from a station to be α. He walks c metres along a slope inclined at an angle β and finds the angle of elevation of the peak of the hill to be ϒ. Show that the height of the peak above the ground is $\frac{c \sin α \sin (γ - β)}{(\sin γ - α)}$ .

Answer:

Suppose, AB is a peak whose height above the ground is t+x.

$In △ D F C, \sin β = \frac{x}{c} \Rightarrow x = c \sin β and \tan β = \frac{x}{y} \Rightarrow y = \frac{x}{\tan β} = \frac{c \sin β}{\sin β} \times \cos β = c \cos β . . . (1) In ∆ A D E, \tan γ = \frac{t}{z} \Rightarrow z = t \cot γ . . . (2) In ∆ A B C, \tan α = \frac{t + x}{y + z} \Rightarrow t + x = (c \cos β \tan α + t c o t γ) \tan α (from (1) a n d (2)) \Rightarrow t - t c o t γ \tan α = c \cos β \tan α - c \sin β (∵ x = c \sin β) \Rightarrow t (1 - \frac{\sin α \cos γ}{\cos α \sin γ}) = c (\frac{\cos β \sin α - \cos α \sin β}{\cos α}) \Rightarrow t (\frac{\sin γ \cos α - \sin α \cos γ}{\cos α \sin γ}) = c \frac{\sin (α - β)}{\cos α} \Rightarrow t \frac{\sin (γ - β)}{\cos α \sin γ} = c \frac{\sin (α - β)}{\cos α} \Rightarrow t = c \frac{\sin γ \sin (α - β)}{\sin (γ - β)} . . . (3) Now, A B = t + x = c \frac{\sin γ \sin (α - β)}{\sin (γ - β)} + c \sin β (using (3)) = c (\frac{\sin γ \sin (α - β)}{\sin (γ - β)} + \sin β) = c [\frac{\sin γ \sin (α - β) + \sin β \sin (γ - β)}{\sin (γ - β)}] = c [\frac{\sin γ \sin α \cos β - \sin β \sin γ \cos α + \sin β \sin γ \cos α - \sin β \cos γ \sin α}{\sin (γ - β)}] = c [\frac{\sin γ \sin α \cos β - \sin β \cos γ \sin α}{\sin (γ - β)}] = c [\frac{\sin α \sin (γ - β)}{\sin (γ - β)}] = \frac{c \sin α \sin (γ - β)}{\sin (γ - β)} Hence proved .$

Page No 10.14:

Question 31:

If the sides a, b and c of âˆ†ABC are in H.P., prove that $\sin^{2} \frac{A}{2}, \sin^{2} \frac{B}{2} and \sin^{2} \frac{C}{2}$ are in H.P.

Answer:

$\sin^{2} \frac{A}{2}, \sin^{2} \frac{B}{2} and \sin^{2} \frac{C}{2} is a H . P . \Leftrightarrow \frac{1}{\sin^{2} \frac{A}{2}}, \frac{1}{\sin^{2} \frac{B}{2}} and \frac{1}{\sin^{2} \frac{C}{2}} is an A . P . \Leftrightarrow \frac{1}{\sin^{2} \frac{B}{2}} - \frac{1}{\sin^{2} \frac{A}{2}} = \frac{1}{\sin^{2} \frac{C}{2}} - \frac{1}{\sin^{2} \frac{B}{2}} \Leftrightarrow \frac{\sin^{2} \frac{A}{2} - \sin^{2} \frac{B}{2}}{\sin^{2} \frac{A}{2} \sin^{2} \frac{B}{2}} = \frac{\sin^{2} \frac{B}{2} - \sin^{2} \frac{C}{2}}{\sin^{2} \frac{B}{2} \sin^{2} \frac{C}{2}} \Leftrightarrow \frac{\sin (\frac{A + B}{2}) \sin (\frac{A - B}{2})}{\sin^{2} \frac{A}{2}} = \frac{\sin (\frac{B + C}{2}) \sin (\frac{B - C}{2})}{\sin^{2} \frac{C}{2}} \Leftrightarrow \frac{\cos (\frac{C}{2}) \sin (\frac{A - B}{2})}{\sin^{2} \frac{A}{2}} = \frac{\cos (\frac{A}{2}) \sin (\frac{B - C}{2})}{\sin^{2} \frac{C}{2}} [As, A + B + C = π] \Leftrightarrow \sin^{2} \frac{C}{2} \cos (\frac{C}{2}) \sin (\frac{A - B}{2}) = \sin^{2} \frac{A}{2} \cos (\frac{A}{2}) \sin (\frac{B - C}{2}) \Leftrightarrow 2 \sin \frac{C}{2} \sin \frac{C}{2} \cos (\frac{C}{2}) \sin (\frac{A - B}{2}) = 2 \sin \frac{A}{2} \sin \frac{A}{2} \cos (\frac{A}{2}) \sin (\frac{B - C}{2}) \Leftrightarrow \sin \frac{C}{2} \sin C \sin (\frac{A - B}{2}) = \sin \frac{A}{2} \sin A \sin (\frac{B - C}{2}) [∵ \sin 2 θ = 2 sinθcosθ] \Leftrightarrow \sin C \cos (\frac{A + B}{2}) \sin (\frac{A - B}{2}) = \sin A \cos (\frac{B + C}{2}) \sin (\frac{B - C}{2}) [As, A + B + C = π] \Leftrightarrow \sin C \frac{(\sin A - \sin B)}{2} = \sin A \frac{(\sin B - \sin C)}{2} [\sin C - \sin D = 2 \cos (\frac{C + D}{2}) \sin (\frac{C - D}{2})] \Leftrightarrow \sin C (\sin A - \sin B) = \sin A (\sin B - \sin C) \Leftrightarrow c k (a k - b k) = a k (b k - c k) (\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k (say)) \Leftrightarrow c a - c b = a b - a c \Leftrightarrow 2 a c = a b + b c \Leftrightarrow \frac{2}{b} = \frac{1}{c} + \frac{1}{a} [multiplying both the sides by a b c] \Leftrightarrow a, b, c are in H . P .$

Page No 10.25:

Question 1:

In $∆ A B C, if a = 5, b = 6 and C = 60 °$ , show that its area is $\frac{15 \sqrt{3}}{2} sq$ . units.

Answer:

$Given : a = 5, b = 6, c = 60 ° Area of a triangle = \frac{1}{2} a b \sin C = \frac{1}{2} \times 5 \times 6 \times \sin 60 ° = 15 \times \frac{\sqrt{3}}{2} sq . units Hence proved .$

Page No 10.25:

Question 2:

In $∆ A B C, if a = \sqrt{2}, b = \sqrt{3} and c = \sqrt{5}$ , show that its area is $\frac{1}{2} \sqrt{6} sq .$ units.

Answer:

$Given: a = \sqrt{2}, b = \sqrt{3}, c = \sqrt{5} ∵ \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \Rightarrow \cos C = \frac{2 + 3 - 5}{2 \times \sqrt{6}} = 0 \Rightarrow \cos C = 0 \Rightarrow \cos C = \cos 90 ° \Rightarrow C = 90 ° Thus, \sin C = \sin 90 ° = 1 Hence, Area of ∆ A B C = \frac{1}{2} a b \sin C = \frac{1}{2} \sqrt{6} \times 1 = \frac{\sqrt{6}}{2} sq . units .$

Page No 10.25:

Question 3:

The sides of a triangle are a = 4, b = 6 and c = 8. Show that $8 \cos A + 16 \cos B + 4 \cos C = 17$ .

Answer:

Given: $a = 4, b = 6 and c = 8 .$
Then,
$\cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{16 + 64 - 36}{2 \times 4 \times 8} = \frac{11}{16} \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{36 + 64 - 16}{2 \times 6 \times 8} = \frac{7}{8} \cos C = \frac{b^{2} + a^{2} - c^{2}}{2 a b} = \frac{16 + 36 - 64}{2 \times 4 \times 6} = \frac{- 1}{4} Now, 8 \cos A + 16 \cos B + 4 \cos C = 8 \times \frac{7}{8} + 16 \times \frac{11}{16} - 4 \times \frac{1}{4} \Rightarrow 8 \cos A + 16 \cos B + 4 \cos C = 7 + 11 - 1 = 17$

Hence proved.

Page No 10.25:

Question 4:

In âˆ† ABC, if a = 18, b = 24 and c = 30, find cos A, cos B and cos C.

Answer:

$GIven : a = 18, b = 24 and c = 30 . \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{576 + 900 - 324}{2 \times 24 \times 30} = \frac{1152}{1140} = \frac{4}{5} \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{324 + 900 - 576}{2 \times 18 \times 30} = \frac{648}{1080} = \frac{3}{5} \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{576 + 324 - 900}{2 \times 24 \times 18} = 0$

Hence, $\cos A = \frac{4}{5}, \cos B = \frac{3}{5}, \cos C = 0$

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Question 5:

In âˆ†ABC, prove the following:
$b (c \cos A - a \cos C) = c^{2} - a^{2}$

Answer:

Let ABC be any triangle.

$Consider b (c \cos A - a \cos C) = b c \cos A - a b \cos C = b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) - a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) = \frac{b^{2} + c^{2} - a^{2} - a^{2} - b^{2} + c^{2}}{2} = \frac{2 (c^{2} - a^{2})}{2} = c^{2} - a^{2}$

Hence proved.

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Question 6:

In âˆ†ABC, prove the following:
$c (a \cos B - b \cos A) = a^{2} - b^{2}$

Answer:

Consider

$c (a \cos B - b \cos A) = c a \cos B - c b \cos A = c a (\frac{a^{2} + c^{2} - b^{2}}{2 a c}) - c b (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) = \frac{a^{2} + c^{2} - b^{2} - b^{2} - c^{2} + a^{2}}{2} = \frac{2 (a^{2} - b^{2})}{2} = a^{2} - b^{2}$

Hence proved.

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Question 7:

In âˆ†ABC, prove the following:
$2 (b c \cos A + c a \cos B + a b \cos C) = a^{2} + b^{2} + c^{2}$

Answer:

LHS = $2 (b c \cos A + c a \cos B + a b \cos C)$

On using the cosine law, we get:
$LHS = 2 [b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) + c a (\frac{a^{2} + c^{2} - b^{2}}{2 a c}) + a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b})]$
$= b^{2} + c^{2} - a^{2} + a^{2} + c^{2} - b^{2} + a^{2} + b^{2} - c^{2} = a^{2} + b^{2} + c^{2} = RHS$

Hence proved.

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Question 8:

In âˆ†ABC, prove the following:
$(c^{2} - a^{2} + b^{2}) \tan A = (a^{2} - b^{2} + c^{2}) \tan B = (b^{2} - c^{2} + a^{2}) \tan C$

Answer:

$We know that \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$

$So, (c^{2} - a^{2} + b^{2}) \tan A = (c^{2} - a^{2} + b^{2}) \frac{\sin A}{\cos A} = (c^{2} - a^{2} + b^{2}) \sin A \frac{2 b c}{b^{2} + c^{2} - a^{2}} = 2 b c \sin A = 2 k a b c . . . (1) (a^{2} - b^{2} + c^{2}) \tan B = (a^{2} - b^{2} + c^{2}) \frac{\sin B}{\cos B} = (a^{2} - b^{2} + c^{2}) \sin B \frac{2 a c}{a^{2} + c^{2} - b^{2}} = 2 a c \sin B = 2 k a b c . . . (2) (b^{2} - a^{2} + c^{2}) \tan C = (b^{2} - a^{2} + c^{2}) \frac{\sin C}{\cos C} = (b^{2} - c^{2} + a^{2}) \sin C \frac{2 a b}{a^{2} + b^{2} - c^{2}} = 2 a b \sin C = 2 k a b c . . . . (3)$

From (1), (2) and (3), we get:
$(c^{2} - a^{2} + b^{2}) \tan A = (a^{2} - b^{2} + c^{2}) \tan B = (b^{2} - c^{2} + a^{2}) \tan C$

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Question 9:

In âˆ†ABC, prove the following:
$\frac{c - b \cos A}{b - c \cos A} = \frac{\cos B}{\cos C}$

Answer:

Let ABC be any triangle.

$LHS = \frac{c - b \cos A}{b - c \cos A} = \frac{a \cos B + b \cos A - b \cos A}{a \cos C + c \cos A - c \cos A} [Using projection formula : c = a \cos B + b \cos A, b = a \cos C + c \cos A] = \frac{a \cos B}{a \cos C} = \frac{\cos B}{\cos C} = RHS$

Hence proved.

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Question 10:

In âˆ†ABC, prove that:
$a (\cos B + \cos C - 1) + b (\cos C + \cos A - 1) + c (\cos A + \cos B - 1) = 0$

Answer:

$Consider the LHS of the given equation . LHS = a (\cos B + \cos C - 1) + b (\cos C + \cos A - 1) + c (\cos A + \cos B - 1) = a \cos B + b \cos C + a \cos C + b \cos A + c \cos A + c \cos B - (a + b + c) = (a \cos B + b \cos A) + (b \cos C + c \cos B) + (a \cos C + c \cos A) - (a + b + c) = c + a + b - (a + b + c) . . (Using projection formula : a = b \cos C + c \cos B, b = a \cos C + c \cos A, c = a \cos B + b \cos A) = 0 = RHS Hence proved .$

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Question 11:

a cos A + b cos B + c cos C = 2b sin A sin C

Answer:

$By sine rule, we know that \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k (say) \Rightarrow a = k \sin A, b = k \sin B, c = k \sin C Now, LHS = a \cos A + b \cos B + c \cos C = k \sin A \cos A + k \sin B \cos B + k \sin C \cos C = \frac{k}{2} (2 \sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C) = \frac{k}{2} (\sin 2 A + \sin 2 B + 2 \sin C \cos C) = \frac{k}{2} (2 \sin \frac{2 A + 2 B}{2} \cos \frac{2 A - 2 B}{2} + 2 \sin C \cos C) = \frac{k}{2} (2 \sin (A + B) \cos (A - B) + 2 \sin C \cos C) = \frac{k}{2} (2 \sin (π - C) \cos (A - B) + 2 \sin C \cos C) (∵ A + B + C = π) = \frac{k}{2} (2 \sin C \cos (A - B) + 2 \sin C \cos C) = \frac{k}{2} \times 2 \sin C (\cos (A - B) + \cos C) = k \sin C (2 \cos (\frac{A - B + C}{2}) \cos (\frac{A - B - C}{2})) = k \sin C (2 \cos (\frac{π - B - B}{2}) \cos (\frac{B + C - A}{2})) (∵ A + B + C = π) = k \sin C (2 \cos (\frac{π - 2 B}{2}) \cos (\frac{π - 2 A}{2})) (∵ A + B + C = π) = k \sin C (2 \cos (\frac{π}{2} - B) \cos (\frac{π}{2} - A)) = 2 k \sin C (\sin B \sin A) = 2 (k \sin B) \sin A \sin C = 2 b \sin A \sin C = RHS ∴ LHS = RHS$

Hence, a cos A + b cos B + c cos C = 2b sin A sin C.

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Question 12:

In âˆ†ABC, prove the following:
$a^{2} = {(b + c)}^{2} - 4 b c \cos^{2} \frac{A}{2}$

Answer:

$RHS = {(b + c)}^{2} - 4 b c \cos^{2} \frac{A}{2} = b^{2} + c^{2} + 2 b c - 4 b c (\frac{1 + \cos A}{2}) = b^{2} + c^{2} + 2 b c - 2 b c (1 + \cos A) = b^{2} + c^{2} + 2 b c (1 - 1 - \cos A) = b^{2} + c^{2} - 2 b c \cos A = b^{2} + c^{2} - 2 b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) (∵ \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}) = b^{2} + c^{2} - b^{2} - c^{2} + a^{2} = a^{2} = LHS$

Hence proved.

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Question 13:

In âˆ†ABC, prove the following:
$4 (b c \cos^{2} \frac{A}{2} + c a \cos^{2} \frac{B}{2} + a b \cos^{2} \frac{C}{2}) = {(a + b + c)}^{2}$

Answer:

$LHS = 4 (b c \cos^{2} \frac{A}{2} + c a \cos^{2} \frac{B}{2} + a b \cos^{2} \frac{C}{2}) = 4 [b c (\frac{1 + \cos A}{2}) + c a (\frac{1 + \cos B}{2}) + a b (\frac{1 + \cos C}{2})] = 2 b c + 2 b c \cos A + 2 c a + 2 c a \cos B + 2 a b + 2 a b \cos C = 2 (a b + b c + c a) + 2 b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) + 2 c a (\frac{c^{2} + a^{2} - b^{2}}{2 c a}) + 2 a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b})$

$= 2 (a b + b c + a c) + b^{2} + c^{2} - a^{2} + c^{2} + a^{2} - b^{2} + a^{2} + b^{2} - c^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c = {(a + b + c)}^{2} = RHS$

Hence proved.

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Question 14:

In âˆ†ABC, prove the following:
$\sin^{3} A \cos (B - C) + \sin^{3} B \cos (C - A) + \sin^{3} C \cos (A - B) = 3 \sin A \sin B \sin C$

Answer:

$Let \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k . . . (1) LHS = \sin^{3} A \cos (B - C) + \sin^{3} B \cos (C - A) + \sin^{3} C \cos (A - B) = \sin^{2} A \{\sin A \cos (B - C)\} + \sin^{2} B \{\sin B \cos (C - A)\} + \sin^{2} A \{\sin A \cos (A - B)\} = \frac{a^{2}}{k^{2}} \{\sin A \cos (B - C)\} + \frac{b^{2}}{k^{2}} \{\sin B \cos (C - A)\} + \frac{c^{2}}{k^{2}} \{\sin A \cos (A - B)\} [from (1)] = \frac{1}{2 k^{2}} [a^{2} \{2 \sin A \cos (B - C)\} + b^{2} \{2 \sin B \cos (C - A)\} + c^{2} \{2 \sin A \cos (A - B)\}] = \frac{1}{2 k^{2}} [a^{2} \{2 \sin (π - (B + C)) \cos (B - C)\} + b^{2} \{2 \sin (π - (A + C)) \cos (C - A)\} + c^{2} \{2 \sin (π - (B + C)) \cos (A - B)\}] = \frac{1}{2 k^{2}} [a^{2} \{2 \sin (B + C) \cos (B - C)\} + b^{2} \{2 \sin (C + A) \cos (C - A)\} + c^{2} \{2 \sin (A + B) \cos (A - B)\}] = \frac{1}{2 k^{2}} [a^{2} \{\sin 2 B + \sin 2 C\} + b^{2} \{\sin 2 C + \sin 2 A\} + c^{2} \{\sin 2 A + \sin 2 B\}] = \frac{1}{2 k^{2}} [2 a^{2} \{\sin B \cos B + \sin C \cos C\} + 2 b^{2} \{\sin C \cos C + \sin A \cos A\} + 2 c^{2} \{\sin A \cos A + \sin B \cos B\}] = \frac{1}{2 k^{3}} [2 a^{2} \{k \sin B \cos B + k \sin C \cos C\} + 2 b^{2} \{k \sin C \cos C + k \sin A \cos A\} + 2 c^{2} \{k \sin A \cos A + k \sin B \cos B\}] = \frac{1}{k^{3}} [a^{2} \{b \cos B + c \cos C\} + 2 b^{2} \{c \cos C + a \cos A\} + 2 c^{2} \{a \cos A + a \cos B\}] = \frac{1}{k^{3}} [a b (a \cos B + b \cos A) + b c (b \cos C + c \cos B) + a c (a \cos C + c \cos A)] = \frac{1}{k^{3}} (a b c + b c a + a c b) = 3 a b c \times \frac{1}{k^{3}} = 3 \sin A \sin B \sin C \times \frac{1}{k^{3}} \times k^{3} = 3 \sin A \sin B \sin C = RHS Hence proved .$

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Question 15:

In $∆ A B C, \frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15}$ . Prove that $\frac{\cos A}{2} = \frac{\cos B}{7} = \frac{\cos C}{11}$ .

Answer:

$Let \frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15} = k \Rightarrow b + c = 12 k, c + a = 13 k, a + b = 15 k \Rightarrow b + c + c + a + a + b = 12 k + 13 k + 15 k \Rightarrow 2 (a + b + c) = 40 k \Rightarrow a + c + b = 20 k \Rightarrow a + 12 k = 20 k (∵ b + c = 12 k) \Rightarrow a = 8 k Also, c + a = 13 k \Rightarrow c = 13 k - a = 13 k - 8 k = 5 k and a + b = 15 k \Rightarrow b = 15 k - a = 15 k - 8 k = 7 k Now, \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{k^{2} (49 + 25 - 64)}{k^{2} (2 \times 35)} = \frac{1}{7} \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{k^{2} (64 + 25 - 49)}{(2 \times 40) k^{2}} = \frac{1}{2} \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{k^{2} (64 + 49 - 25)}{(2 \times 56) k^{2}} = \frac{11}{14} ∴ \cos A : \cos B : \cos C = \frac{1}{7} : \frac{1}{2} : \frac{11}{14} = 2 : 7 : 11 \Rightarrow \frac{\cos A}{2} = \frac{\cos B}{7} = \frac{\cos C}{11}$

Hence proved.

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Question 16:

In $∆ A B C, if ∠ B = 60 °,$ prove that $(a + b + c) (a - b + c) = 3 c a$ .

Answer:

$Given, ∠ B = 60 ° We know that, \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} \Rightarrow \cos 60 ° = \frac{a^{2} + c^{2} - b^{2}}{2 a c} \Rightarrow \frac{1}{2} = \frac{a^{2} + c^{2} - b^{2}}{2 a c} (∵ \cos 60 ° = \frac{1}{2}) \Rightarrow a c = a^{2} + c^{2} - b^{2} \Rightarrow 3 a c - 2 a c = a^{2} + c^{2} - b^{2} \Rightarrow 3 a c = a^{2} + c^{2} - b^{2} + 2 a c \Rightarrow 3 a c = a^{2} + c^{2} + 2 a c - b^{2} \Rightarrow 3 a c = {(a + c)}^{2} - b^{2} \Rightarrow 3 a c = (a + c + b) (a + c - b) \Rightarrow 3 a c = (a + b + c) (a - b + c)$

Hence proved.

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Question 17:

If in $∆ A B C, \cos^{2} A + \cos^{2} B + \cos^{2} C = 1$ , prove that the triangle is right-angled.

Answer:

Let ABC be any triangle.
In $∆ ABC$ ,
$\cos^{2} A + \cos^{2} B + \cos^{2} C = 1 \Rightarrow \cos^{2} A + \cos^{2} B + \cos^{2} [π - (B + A)] = 1 (∵ A + B + C = π) \Rightarrow \cos^{2} A + \cos^{2} B + \cos^{2} (B + A) = 1 \Rightarrow \cos^{2} A + \cos^{2} B = 1 - \cos^{2} (B + A) \Rightarrow \cos^{2} A + \cos^{2} B = \sin^{2} (B + A) \Rightarrow \cos^{2} A + \cos^{2} B = {(\sin A \cos B + \cos A \sin B)}^{2} \Rightarrow \cos^{2} A + \cos^{2} B = \sin^{2} A \cos^{2} B + \cos^{2} A \sin^{2} B + 2 \sin A \sin B \cos A \cos B \Rightarrow \cos^{2} A (1 - \sin^{2} B) + \cos^{2} B (1 - \sin^{2} A) = 2 \sin A \sin B \cos A \cos B \Rightarrow 2 \cos^{2} A \cos^{2} B = 2 \sin A \sin B \cos A \cos B \Rightarrow \cos A \cos B = \sin A \sin B \Rightarrow \cos A \cos B - \sin A \sin B = 0 \Rightarrow \cos (A + B) = 0 \Rightarrow \cos (A + B) = \cos 90^{°} \Rightarrow A + B = 90^{°} \Rightarrow C = 90^{°} (∵ A + B + C = 180^{°})$

Hence, $∆$ ABC is right angled.

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Question 18:

In $∆ A B C if \cos C = \frac{\sin A}{2 \sin B}$ , prove that the triangle is isosceles.

Answer:

Let $∆ A B C$ be any triangle.

Suppose $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k$
If $\cos C = \frac{\sin A}{2 \sin B}$ , then

$\frac{b^{2} + a^{2} - c^{2}}{2 a b} = \frac{k a}{2 k b} (∵ \cos C = \frac{b^{2} + a^{2} - c^{2}}{2 a b})$

^{$\Rightarrow b^{2} + a^{2} - c^{2} = a^{2} \Rightarrow b^{2} - c^{2} = 0 \Rightarrow (b - c) (b + c) = 0 \Rightarrow b - c = 0 \Rightarrow b = c (∵ b, c > 0)$}

Thus, the lengths of two sides of the $∆ A B C$ are equal.

Hence, $∆ ABC$ is an isosceles triangle.

Page No 10.26:

Question 19:

Two ships leave a port at the same time. One goes 24 km/hr in the direction N 38° E and other travels 32 km/hr in the direction S 52° E. Find the distance between the ships at the end of 3 hrs.

Answer:

$After three hours, let the ships be at P and Q respectively . Then, O P = 24 \times 3 = 72 km and O Q = 32 \times 3 = 96 km From figure, we have ∠ P O Q = 180 ° - ∠ N O P - ∠ S O Q = 180 ° - 38 ° - 52 ° = 90 ° Now, Since O P Q is a right angled triangle ∴ P Q^{2} = O P^{2} + O Q^{2} \Rightarrow P Q^{2} = 72^{2} + 96^{2} \Rightarrow P Q^{2} = 5184 + 9216 \Rightarrow P Q^{2} = 14400 \Rightarrow P Q = \sqrt{14400} = 120 km Hence, the distance between the ships after 3 hours is 120 km .$

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Question 1:

Mark the correct alternative in each of the following:

In any âˆ†ABC, $\sum_{}^{} a^{2} (\sin B - \sin C) =$

(a) $a^{2} + b^{2} + c^{2}$ (b) $a^{2}$ (c) $b^{2}$ (d) 0

Answer:

Using sine rule, we have

$\sum_{}^{} a^{2} (\sin B - \sin C)$

$= a^{2} (\frac{b}{k} - \frac{c}{k}) + b^{2} (\frac{c}{k} - \frac{a}{k}) + c^{2} (\frac{a}{k} - \frac{b}{k}) = \frac{1}{k} (a^{2} b - a^{2} c + b^{2} c - b^{2} a + c^{2} a - c^{2} b)$

This expression cannot be simplified to match with any of the given options.

However, if the quesion is "In any âˆ†ABC, $\sum_{}^{} a^{2} (\sin^{2} B - \sin^{2} C) =$ ", then the solution is as follows.

Using sine rule, we have

$\sum_{}^{} a^{2} (\sin^{2} B - \sin^{2} C)$

$= a^{2} (\frac{b^{2}}{k^{2}} - \frac{c^{2}}{k^{2}}) + b^{2} (\frac{c^{2}}{k^{2}} - \frac{a^{2}}{k^{2}}) + c^{2} (\frac{a^{2}}{k^{2}} - \frac{b^{2}}{k^{2}}) = \frac{1}{k^{2}} (a^{2} b^{2} - a^{2} c^{2} + b^{2} c^{2} - b^{2} a^{2} + c^{2} a^{2} - c^{2} b^{2}) = \frac{1}{k^{2}} \times 0 = 0$

Hence, the correct answer is option (d).

Disclaimer: The question given in the book in incorrect or there is some printing mistake in the question.

Page No 10.26:

Question 2:

Mark the correct alternative in each of the following:

In a âˆ†ABC, if a = 2, $∠ B = 60 °$ and $∠ C = 75 °$ , then b =

(a) $\sqrt{3}$ (b) $\sqrt{6}$ (c) $\sqrt{9}$ (d) $1 + \sqrt{2}$

Answer:

It is given that a = 2, $∠ B = 60 °$ and $∠ C = 75 °$ .

In âˆ†ABC,

$∠ A + ∠ B + ∠ C = 180 ° (Angle sum property) \Rightarrow ∠ A + 60 ° + 75 ° = 180 ° \Rightarrow ∠ A = 180 ° - 135 ° = 45 °$

Using sine rule, we get

$\frac{2}{\sin 45 °} = \frac{b}{\sin 60 °} (\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}) \Rightarrow b = \frac{2 \times \frac{\sqrt{3}}{2}}{\frac{1}{\sqrt{2}}} = \sqrt{6}$

Hence, the correct answer is option (b).

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Question 3:

Mark the correct alternative in each of the following:

If the sides of a triangle are in the ratio $1 : \sqrt{3} : 2$ , then the measure of its greatest angle is

(a) $\frac{π}{6}$ (b) $\frac{π}{3}$ (c) $\frac{π}{2}$ (d) $\frac{2 π}{3}$

Answer:

Let âˆ†ABC be the given triangle such that its sides are in the ratio $1 : \sqrt{3} : 2$ .

$∴ a = k, b = \sqrt{3} k, c = 2 k$

Now, $a^{2} + b^{2} = k^{2} + 3 k^{2} = 4 k^{2} = c^{2}$

So, âˆ†ABC is a right triangle right angled at C.

$∴ C = 90 °$

Using sine rule, we have

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow \frac{k}{\sin A} = \frac{\sqrt{3} k}{\sin B} = \frac{2 k}{\sin 90 °} \Rightarrow \sin A = \frac{1}{2} and \sin B = \frac{\sqrt{3}}{2} \Rightarrow A = 30 ° and B = 60 °$

Thus, the measure of its greatest angle is $\frac{π}{2}$ .

Hence, the correct answer is option (c).

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Question 4:

Mark the correct alternative in each of the following:

In any âˆ†ABC, 2(bc cosA + ca cosB + ab cosC) =

(a) $a b c$ (b) $a + b + c$ (c) $a^{2} + b^{2} + c^{2}$ (d) $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}$

Answer:

Using cosine rule, we have

$2 (b c \cos A + c a \cos B + a b \cos C) = 2 b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) + 2 c a (\frac{c^{2} + a^{2} - b^{2}}{2 c a}) + 2 a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) = b^{2} + c^{2} - a^{2} + c^{2} + a^{2} - b^{2} + a^{2} + b^{2} - c^{2} = a^{2} + b^{2} + c^{2}$

Hence, the correct answer is option (c).

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Question 5:

Mark the correct alternative in each of the following:

In a triangle ABC, a = 4, b = 3, $∠ A = 60 °$ then c is a root of the equation

(a) $c^{2} - 3 c - 7 = 0$ (b) $c^{2} + 3 c + 7 = 0$ (c) $c^{2} - 3 c + 7 = 0$ (d) $c^{2} + 3 c - 7 = 0$

Answer:

It is given that a = 4, b = 3 and $∠ A = 60 °$ .

Using cosine rule, we have

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow \cos 60 ° = \frac{9 + c^{2} - 16}{2 \times 3 \times c} \Rightarrow \frac{1}{2} = \frac{c^{2} - 7}{6 c} \Rightarrow c^{2} - 7 = 3 c \Rightarrow c^{2} - 3 c - 7 = 0$

Thus, c is the root of $c^{2} - 3 c - 7 = 0$ .

Hence, the correct answer is option (a).

Page No 10.26:

Question 6:

Mark the correct alternative in each of the following:

In a âˆ†ABC, if $(c + a + b) (a + b - c) = a b$ , then the measure of angle C is

(a) $\frac{π}{3}$ (b) $\frac{π}{6}$ (c) $\frac{2 π}{3}$ (d) $\frac{π}{2}$

Answer:

Given: $(c + a + b) (a + b - c) = a b$

$\Rightarrow {(a + b)}^{2} - c^{2} = a b \Rightarrow a^{2} + b^{2} + 2 a b - c^{2} = a b \Rightarrow a^{2} + b^{2} - c^{2} = - a b \Rightarrow \frac{a^{2} + b^{2} - c^{2}}{2 a b} = - \frac{1}{2}$
$\Rightarrow \cos C = - \frac{1}{2} = \cos \frac{2 π}{3} (Using cosine rule) \Rightarrow C = \frac{2 π}{3}$
Thus, the measure of angle C is $\frac{2 π}{3}$ .

Hence, the correct answer is option (c).

Page No 10.26:

Question 7:

Mark the correct alternative in each of the following:

In any âˆ†ABC, the value of $2 a c \sin (\frac{A - B + C}{2})$ is

(a) $a^{2} + b^{2} - c^{2}$ (b) $c^{2} + a^{2} - b^{2}$ (c) $b^{2} - c^{2} - a^{2}$ (d) $c^{2} - a^{2} - b^{2}$

Answer:

In âˆ†ABC,

$A + B + C = π (Angle sum property) \Rightarrow A + C = π - B$

$∴ 2 a c \sin (\frac{A - B + C}{2}) = 2 a c \sin (\frac{π - 2 B}{2}) = 2 a c \sin (\frac{π}{2} - B) = 2 a c \cos B$
$= 2 a c (\frac{c^{2} + a^{2} - b^{2}}{2 c a}) (Using cosine rule) = c^{2} + a^{2} - b^{2}$

Hence, the correct answer is option (b).

Page No 10.26:

Question 8:

Mark the correct alternative in each of the following:

In any âˆ†ABC, $a (b \cos C - c \cos B) =$

(a) $a^{2}$ (b) $b^{2} - c^{2}$ (c) 0 (d) $b^{2} + c^{2}$

Answer:

Using cosine rule, we have

$a (b \cos C - c \cos B) = a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) - c a (\frac{c^{2} + a^{2} - b^{2}}{2 c a}) = \frac{a^{2} + b^{2} - c^{2} - c^{2} - a^{2} + b^{2}}{2} = \frac{2 b^{2} - 2 c^{2}}{2} = b^{2} - c^{2}$

Hence, the correct answer is option (b).

Page No 10.26:

Question 1:

In a âˆ†ABC, if $\frac{\sin (A - B)}{\sin (A + B)} = \frac{a^{2} - b^{2}}{k},$ then k = ___________.

Answer:

In âˆ†ABC,

Given $\frac{\sin (A - B)}{\sin (A + B)} = \frac{\sin A \cos B - \cos A \sin B}{\sin A \cos B + \cos A \sin B}$

$Since \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} and \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = x say$
$\begin{array}{rcl} i . e \frac{\sin (A - B)}{\sin (A + B)} & = & \frac{(\frac{a}{x}) (\frac{a^{2} + c^{2} - b^{2}}{2 a c}) - (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) (\frac{b}{x})}{\frac{a}{x} (\frac{a^{2} + c^{2} - b^{2}}{2 a c}) + (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) (\frac{b}{x})} \\ = & \frac{\frac{a^{2} + c^{2} - b^{2}}{2 c x} - \frac{b^{2} + c^{2} - a^{2}}{2 c x}}{\frac{a^{2} + c^{2} - b^{2}}{2 c x} + \frac{b^{2} + c^{2} - a^{2}}{2 c x}} \\ = & \frac{a^{2} + c^{2} - b^{2} - b^{2} - c^{2} + a^{2}}{a^{2} + c^{2} - b^{2} + b^{2} + c^{2} - a^{2}} \\ = & \frac{2 a^{2} - 2 b^{2}}{2 c^{2}} = \frac{a^{2} - b^{2}}{c^{2}} = \frac{a^{2} - b^{2}}{k} (given) \\ i . e . k & = & c^{2} \end{array}$

Page No 10.26:

Question 2:

In a âˆ†ABC, if c² + a² – b² = ac, then the measure of angle B is ____________.

Answer:

In a âˆ†ABC,
if c² + a² − b²= ac

$\begin{array}{rcl} Since \cos B & = & \frac{c^{2} + a^{2} - b^{2}}{2 a c} \\ = & \frac{a c}{2 a c} \end{array}$

i.e cosB = $\frac{1}{2}$

i.e B = $\frac{π}{3}$

Page No 10.27:

Question 3:

In a âˆ†ABC, if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ and a = 2, then area of âˆ†ABC is equal to __________.

Answer:

In a âˆ†ABC

Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c} . . . (1) and a = 2$
Using sine formula
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} . . . (2)$

using above two, (1) and (2)
we get,
$\frac{\cos A}{\sin A} = \frac{\cos B}{\sin B} = \frac{\cos C}{\sin C}$

⇒ cotA = cotB = cotC
⇒ angle A = B = C = 60^°
⇒ âˆ†ABC is an equilateral triangle

$\begin{array}{rcl} ∴ Area of triangle ABC & = & \frac{\sqrt{3}}{4} a^{2} \\ = & \frac{\sqrt{3}}{4} \times 4 \\ = & \sqrt{3} {unit}^{2} \end{array}$
$o r \sqrt{3} Sq . unit$

Page No 10.27:

Question 4:

In a triangle ABC, if a = 2, b = 4 and $A + B = \frac{2 π}{3},$ then area of âˆ†ABC is __________.

Answer:

Since A + B = $\frac{2 π}{3}$ given

and A + B + C = $π$ (Angle sum property)

$\begin{array}{rcl} \Rightarrow & C = π - \frac{2 π}{3} \\ C & = & \frac{π}{3} \end{array}$
Since area of triangle ABC = $\frac{1}{2}$ ab sinC
$= \frac{1}{2} \times 2 \times 4 \times \sin \frac{π}{3} (∴ a = 2 and b = 4 given) = 4 \times \frac{\sqrt{3}}{2} = 3 \sqrt{3} sq . units$

Page No 10.27:

Question 5:

The angles A, B, C of a âˆ†ABC are in AP and the sides a, b, c are in G.P. If a² + c² = λb², then λ = ____________.

Answer:

Since A, B, C are A.P
⇒ 2B = A + C
Since A + B + C = $π$ (By angle sum property)
⇒ 3B = $π$
$i . e B = \frac{π}{3} . . . (1)$
also,
Since a, b, c are in g.p
⇒ b² = ac ....(2)
Using cosB = $\frac{a^{2} + c^{2} - b^{2}}{2 a c}$

i.e $\cos \frac{π}{3} = \frac{a^{2} + c^{2} - a c}{2 a c}$ from (1) and (2).

$\Rightarrow \frac{1}{2} = \frac{a^{2} + c^{2} - a c}{2 a c}$
⇒ ac = a² + c² - ac
⇒ a² + c² = 2ac
⇒ a² + c² = 2b² from (2)
Hence $λ$ = 2

Page No 10.27:

Question 6:

In a âˆ†ABC, if ∠C = 60°, a = 47 cm and b = 94 cm, then c² = ____________.

Answer:

Given ∠C = 60^âˆ˜ , a = 47 , b = 94 In âˆ†ABC

Using cosC = $\frac{a^{2} + b^{2} - c^{2}}{2 a b}$

i.e cos 60^âˆ˜ = $\frac{{(47)}^{2} + {(94)}^{2} - C^{2}}{2 \times 47 \times 94}$

i.e $2 \times \frac{1}{2} \times 47 \times 94 = {(47)}^{2} + {(94)}^{2} - C^{2}$

i.e C² = (47)² + (94)² − 47 × 47 × 2
= − (47)² + (94)²
∴ C² = 8836 − 22049 = 6627

Page No 10.27:

Question 7:

In a âˆ†ABC, if $∠ C = \frac{π}{2}, ∠ A = \frac{π}{6},$ c = 20, then a = ____________.

Answer:

In âˆ†ABC

∠C = $\frac{π}{2}$ , ∠ A = $\frac{π}{6}$
⇒∠B = $\frac{π}{3}$ (By angle sum property A + B + C = $π$ )
using sina formula,

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

⇒ $\frac{a}{\sin \frac{π}{6}} = \frac{20}{s in \frac{π}{2}}$ ( $∵$ c = 20 given)
⇒ a = sin $\frac{π}{6} \times 20$ ( $∵$ sin π/2 = 1)
$\Rightarrow a = \frac{1}{2} \times 20$
i.e a = 10

Page No 10.27:

Question 8:

In a âˆ†ABC, if $\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = k (a^{2} + b^{2} + c^{2}),$ then k = ___________________.

Answer:

Using cosine formula

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, \cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} and \cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} L . H . S = \frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \times \frac{1}{a} + \frac{a^{2} + c^{2} - b^{2}}{2 a c} \times \frac{1}{b} + \frac{a^{2} + b^{2} - c^{2}}{2 a b} \times \frac{1}{c} = \frac{b^{2} + c^{2} - a^{2} + a^{2} + c^{2} - b^{2} + a^{2} + b^{2} - c^{2}}{2 a b c} = \frac{a^{2} + b^{2} + c^{2}}{2 a b c} = k (a^{2} + b^{2} + c^{2}) = R . H . S (given) \Rightarrow k = \frac{1}{2 a b c}$

Page No 10.27:

Question 9:

In a âˆ†ABC, if c² sin A sin B = ab, then A + B = ______________.

Answer:

Since c² sinA sinB = ab

$\Rightarrow c^{2} = \frac{a b}{\sin A \sin B} = \frac{a}{\sin A} \times \frac{b}{\sin B} c^{2} = {(\frac{a}{\sin A})}^{2} (since by sine rule \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}) \Rightarrow c = \frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow a = c \sin A and b = c \sin B i . e \sin A = \frac{a}{c} and \sin B = \frac{b}{c} \Rightarrow ∆ is a right angle triangle at C \Rightarrow A + B = \frac{π}{2}$

Page No 10.27:

Question 10:

In a âˆ†ABC, if a = 8, b = 9 and 3 cos C = 2, then C = ____________.

Answer:

a = 8 , b = 9
3 cosC = 2

⇒ cosC = $\frac{2}{3}$
also using, cosine formula

$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} \Rightarrow \frac{2}{3} = \frac{8^{2} + 9^{2} - c^{2}}{2 \times 8 \times 9} \Rightarrow 4 \times 8 \times 3 = 8^{2} + 9^{2} - c^{2} \Rightarrow C^{2} = 8^{2} + 9^{2} - 12 \times 8 = 49 i . e C = 7$

Page No 10.27:

Question 11:

In a âˆ†ABC, if $b = \sqrt{3}, c = 1$ and $B - C = \frac{π}{2},$ then A = ______________.

Answer:

Given b = $\sqrt{3}$ , c = 1 and B – C = $\frac{π}{2}$ In a triangle ABC

By angle sum property,

Since A + B + C = $\frac{π}{2}$ ....(1)

and by Sine formula,

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow \frac{b}{\sin B} = \frac{c}{\sin C} \Rightarrow \frac{\sqrt{3}}{\sin (\frac{π}{2} + c)} = \frac{c = 1}{\sin C}$
$i . e \frac{\sqrt{3}}{\cos C} = \frac{1}{\sin C} (∵ \sin (\frac{π}{2} + c) = \cos C) \Rightarrow \frac{\sin C}{\cos C} = \frac{1}{\sqrt{3}} \Rightarrow t an C = \frac{1}{\sqrt{3}} i . e C = 30^{°} = \frac{π}{6} \Rightarrow B = \frac{π}{2} + \frac{π}{6} = \frac{4 π}{6} \Rightarrow A = π - B - C (from (1)) = π - \frac{2 π}{3} - \frac{π}{6} = \frac{6 π - 4 π - π}{6} i . e A = \frac{π}{6}$

Page No 10.27:

Question 12:

If angles of a triangle are in A.P. and $b : c = \sqrt{3} : \sqrt{2},$ then C = ______________.

Answer:

If angle of a triangle ABC are in A.P
⇒ 2∠B = ∠A + ∠C

and $\frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}}$
By angle sum property
∠A + ∠B + ∠C = $π$
⇒ 2∠B + ∠B = $π$
⇒ ∠B = $\frac{π}{3}$
also,
Using Sine formula
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} i . e \frac{b}{\sin B} = \frac{c}{\sin C} i . e \frac{b}{c} = \frac{\sin B}{\sin C} i . e \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin \frac{π}{3}}{\sin C} (∵ \frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}} given) i . e \sin C = \frac{\sqrt{3}}{2} \times \frac{\sqrt{2}}{\sqrt{3}} i . e \sin C = \frac{1}{\sqrt{2}} i . e \sin C = \frac{π}{4}$

Page No 10.27:

Question 13:

If the sides of a âˆ†ABC are $a, b, \sqrt{a^{2} + a b + b^{2}},$ then the measure of the largest angle is ____________.

Answer:

Let us suppose the greatest angle is c
Using cosine formula,
$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} Since C = \sqrt{a^{2} + a b + b^{2}} \Rightarrow C^{2} = a^{2} + a b + b^{2} ie \cos C = \frac{a^{2} + b^{2} - a^{2} - a b - b^{2}}{2 a b} = \frac{- 1}{2} ie \cos C = \frac{1}{2} ie C = \frac{2 π}{3}$

Page No 10.27:

Question 14:

In a âˆ†ABC, if a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c², then B = __________.

Answer:

In a âˆ†ABC,
Given a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c²
i.e a⁴ + b⁴ − 2a²b² + c²− 2b²c² = 0
i.e (a² + c²)² + b⁴− 2a²b² − 2b²c² − 2a²c² = 0
i.e (a² + c²)² + b⁴ − 2b² (a² + c²) − 2a²c² = 0
i.e (a² + c² − b²)²= 2a²c²
i.e a² + c² − b² = $\sqrt{2} a c$
Using cosine formula,
i.e cosB = $\frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{\sqrt{2} a c}{2 a c}$
cosB = $\frac{1}{\sqrt{2}}$
i.e $B = \frac{π}{4} or π - \frac{π}{4} = \frac{3 π}{4}$

Page No 10.27:

Question 15:

In a âˆ†ABC, if a = 4, b = 3, $A = \frac{π}{3} .$ Then side C is given by ____________.

Answer:

In âˆ†ABC
if a = 4, b = 3, A = $\frac{π}{3}$

$Since \cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} i . e \cos \frac{π}{3} = \frac{9 + c^{2} - 16}{2 \times 3 \times c} i . e \frac{1}{2} = \frac{c^{2} - 7}{6 c} i . e 3 c = c^{2} - 7 i . e c^{2} - 3 c - 7 = 0$

Page No 10.27:

Question 1:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

Find the area of the triangle âˆ†ABC in which a = 1, b = 2 and $∠ C = 60 °$ .

Answer:

In âˆ†ABC, a = 1, b = 2 and $∠ C = 60 °$ .

∴ Area of the âˆ†ABC

$= \frac{1}{2} a b \sin C = \frac{1}{2} \times 1 \times 2 \times \sin 60 ° = \frac{1}{2} \times 2 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} square units$

Page No 10.27:

Question 2:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if b = $\sqrt{3}$ , c = 1 and $∠ A = 30 °$ , find a.

Answer:

In âˆ†ABC, b = $\sqrt{3}$ , c = 1 and $∠ A = 30 °$ .

Using cosine formula, we have

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow \cos 30 ° = \frac{{(\sqrt{3})}^{2} + {(1)}^{2} - a^{2}}{2 \times \sqrt{3} \times 1} \Rightarrow \frac{\sqrt{3}}{2} = \frac{4 - a^{2}}{2 \sqrt{3}} \Rightarrow 3 = 4 - a^{2} \Rightarrow a^{2} = 4 - 3 = 1 \Rightarrow a = 1$

Page No 10.27:

Question 3:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if $\cos A = \frac{\sin B}{2 \sin C}$ , then show that c = a.

Answer:

Given: $\cos A = \frac{\sin B}{2 \sin C}$

$\Rightarrow \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{b}{2 c}$ (Using sine rule and cosine rule)

$\Rightarrow b^{2} + c^{2} - a^{2} = b^{2}$

$\Rightarrow c^{2} = a^{2}$

$\Rightarrow c = a$

Page No 10.27:

Question 4:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if b = 20, c = 21 and $\sin A = \frac{3}{5}$ , find a.

Answer:

In âˆ†ABC, b = 20, c = 21 and $\sin A = \frac{3}{5}$ .

Using cosine rule, we have

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow \sqrt{1 - {(\frac{3}{5})}^{2}} = \frac{20^{2} + 21^{2} - a^{2}}{2 \times 20 \times 21} (\cos^{2} A + \sin^{2} A = 1) \Rightarrow \sqrt{\frac{16}{25}} = \frac{400 + 441 - a^{2}}{840} \Rightarrow \frac{4}{5} = \frac{841 - a^{2}}{840} \Rightarrow 672 = 841 - a^{2}$

$\Rightarrow a^{2} = 841 - 672 = 169 \Rightarrow a = 13$

Page No 10.27:

Question 5:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In a âˆ†ABC, if sinA and sinB are the roots of the equation $c^{2} x^{2} - c (a + b) x + a b = 0$ , then find $∠ C$ .

Answer:

It is given that sinA and sinB are the roots of the equation $c^{2} x^{2} - c (a + b) x + a b = 0$ .

$∴ \sin A + \sin B = - \frac{- c (a + b)}{c^{2}} (Sum of roots = - \frac{b}{a}) \Rightarrow \sin A + \sin B = \frac{a + b}{c} \Rightarrow \sin A + \sin B = \frac{k \sin A + k \sin B}{k \sin C} (Sine rule)$
$\Rightarrow \sin A + \sin B = \frac{\sin A + \sin B}{\sin C} \Rightarrow \sin C = 1 = \sin 90 ° \Rightarrow C = 90 °$

Page No 10.27:

Question 6:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In âˆ†ABC, if a = 8, b = 10, c = 12 and C = λA, find the value of λ.

Answer:

Using cosine rule, we have

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} \Rightarrow \cos A = \frac{10^{2} + 12^{2} - 8^{2}}{2 \times 10 \times 12} \Rightarrow \cos A = \frac{100 + 144 - 64}{240} \Rightarrow \cos A = \frac{180}{240} = \frac{3}{4} . . . . . (1)$

Now, using sine rule, we have

$\frac{a}{\sin A} = \frac{c}{\sin C} \Rightarrow \frac{8}{\sin A} = \frac{12}{\sin λ A} \Rightarrow \sin λ A = \frac{3}{2} \sin A \Rightarrow \sin λ A = 2 \times \frac{3}{4} \sin A$
$\Rightarrow \sin λ A = 2 \sin A \cos A [Using (1)] \Rightarrow \sin λ A = \sin 2 A \Rightarrow λ = 2$

Page No 10.28:

Question 7:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

If the sides of a triangle are proportional to 2, $\sqrt{6}$ and $\sqrt{3} - 1$ , find the measure of its greatest angle.

Answer:

Let âˆ†ABC be the triangle such that a = 2, b = $\sqrt{6}$ and c = $\sqrt{3} - 1$ .

Clearly, b > a > c. Then,

$∠$ B is the greatest angle of âˆ†ABC. (Greatest side has greatest angle opposite to it)

Using cosine formula, we have

$\cos B = \frac{c^{2} + a^{2} - b^{2}}{2 c a} \Rightarrow \cos B = \frac{{(\sqrt{3} - 1)}^{2} + 2^{2} - {(\sqrt{6})}^{2}}{2 \times (\sqrt{3} - 1) \times 2} \Rightarrow \cos B = \frac{3 + 1 - 2 \sqrt{3} + 4 - 6}{4 (\sqrt{3} - 1)}$
$\Rightarrow \cos B = \frac{2 - 2 \sqrt{3}}{4 (\sqrt{3} - 1)} = \frac{- 2 (\sqrt{3} - 1)}{4 (\sqrt{3} - 1)} \Rightarrow \cos B = - \frac{1}{2} = \cos 120 ° \Rightarrow B = 120 °$

Hence, the measure of its greatest angle is 120º.

Page No 10.28:

Question 8:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

If in a âˆ†ABC, $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$ , then find the measures of angles A, B, C.

Answer:

In âˆ†ABC,

$\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c} \Rightarrow \frac{\cos A}{k \sin A} = \frac{\cos B}{k \sin B} = \frac{\cos C}{k \sin C} (Using sine rule) \Rightarrow \cot A = \cot B = \cot C \Rightarrow A = B = C$

⇒ âˆ†ABC is an equilateral triangle.

∴ A = B = C = 60º

Page No 10.28:

Question 9:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In any triangle ABC, find the value of $a \sin (B - C) + b \sin (C - A) + c \sin (A - B)$ .

Answer:

Using sine rule, we have

$a \sin (B - C) + b \sin (C - A) + c \sin (A - B) = k \sin A \sin (B - C) + k \sin B \sin (C - A) + k \sin C \sin (A - B) = k \sin [π - (B + C)] \sin (B - C) + k \sin [π - (C + A)] \sin (C - A) + k \sin [π - (A + B)] \sin (A - B)$
$= k [\sin (B + C) \sin (B - C) + \sin (C + A) \sin (C - A) + \sin (A + B) \sin (A - B)] = k (\sin^{2} B - \sin^{2} C + \sin^{2} C - \sin^{2} A + \sin^{2} A - \sin^{2} B) = k \times 0 = 0$

Hence, the required value is 0.

Page No 10.28:

Question 10:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question.

In any âˆ†ABC, find the value of $\sum_{}^{} a (\sin B - \sin C)$ .

Answer:

Using sine rule, we have

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \Rightarrow a = k \sin A, b = k \sin B, c = k \sin C$

$∴ \sum_{}^{} a (\sin B - \sin C) = \sum_{}^{} k \sin A (\sin B - \sin C)$

$= k \sum_{}^{} \sin A (\sin B - \sin C)$

$= k [\sin A (\sin B - \sin C) + \sin B (\sin C - \sin A) + \sin C (\sin A - \sin B)]$

$= k (\sin A \sin B - \sin A \sin C + \sin B \sin C - \sin B \sin A + \sin C \sin A - \sin C \sin B)$

$= k \times 0 = 0$

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