RD Sharma 2022 Mcqs Solutions for Class 10 Maths Chapter 10 - Trigonometric Ratios

Share with your friends

RD Sharma 2022 Mcqs Solutions for Class 10 Maths Chapter 10 Trigonometric Ratios are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios are extremely popular among class 10 students for Maths Trigonometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2022 Mcqs Book of class 10 Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2022 Mcqs Solutions. All RD Sharma 2022 Mcqs Solutions for class 10 Maths are prepared by experts and are 100% accurate.

Page No 155:

Question 1:

If sinθ = x and secθ = y, then tanθ is equal to
â€‹
(a) xy

(b) $\frac{x}{y}$

(c) $\frac{y}{x}$

(d) $\frac{1}{x y}$

Answer:

Given: sinθ = x and secθ = y
$\Rightarrow \frac{1}{\cos θ} = y \Rightarrow \cos θ = \frac{1}{y}$

$\begin{array}{rcl} ∴ \tan θ & = & \frac{\sin θ}{\cos θ} \\ = & \frac{x}{\frac{1}{y}} \\ = & x y \end{array}$

Hence, the correct answer is option (a).

Page No 155:

Question 2:

Given that $sinθ = \frac{a}{b}$ , then tanθ is equal to
â€‹
(a) $\frac{b}{\sqrt{a^{2} + b^{2}}}$ â€‹

(b) $\frac{b}{\sqrt{b^{2} - a^{2}}}$

(c) $\frac{a}{\sqrt{a^{2} - b^{2}}}$

(d) $\frac{a}{\sqrt{b^{2} - a^{2}}}$

Answer:

Given : $\sin θ = \frac{a}{b}$
We know that, sin²θ + cos²θ =1
⇒ cos²θ = 1 $-$ sin²θ
$\begin{array}{rcl} \Rightarrow & \cos θ = \sqrt{1 - \sin^{2} θ} \\ \Rightarrow & \cos θ = \sqrt{1 - {(\frac{a}{b})}^{2}} \\ \Rightarrow & \cos θ = \frac{\sqrt{b^{2} - a^{2}}}{b} \end{array}$

$\begin{array}{rcl} ∴ \tan θ & = & \frac{\sin θ}{\cos θ} \\ = & \frac{\frac{a}{b}}{\frac{\sqrt{b^{2} - a^{2}}}{b}} \\ = & \frac{a}{\sqrt{b^{2} - a^{2}}} \end{array}$

Hence, the correct answer is option (d).

Page No 155:

Question 3:

If 4 tanβ = 3, then $\frac{4 \sin β - 3 \cos β}{4 \sin β + 3 \cos β} =$
â€‹
(a) 0â€‹

(b) $\frac{1}{3}$

(c) $\frac{2}{3}$

(d) $\frac{7}{25}$

Answer:

Given: 4 tan β = 3 .....(1)

We have to evaluate, $\frac{4 \sin β - 3 \cos β}{4 \sin β + 3 \cos β}$

Dividing numerator and denominator by cos β and substituting (1), we get

$\begin{array}{rcl} \frac{\frac{4 \sin β - 3 \cos β}{\cos β}}{\frac{4 \sin β + 3 \cos β}{\cos β}} & = & \frac{\frac{4 \sin β}{\cos β} - 3 \frac{\cos β}{\cos β}}{\frac{4 \sin β}{\cos β} + 3 \frac{\cos β}{\cos β}} \\ = & \frac{4 \tan β - 3}{4 \tan β + 3} \\ = & \frac{3 - 3}{3 + 3} \\ = & \frac{0}{6} \\ = & 0 \end{array}$

Hence, the correct answer is option (a).

Page No 155:

Question 4:

If ΔABC right angled at B. If $tanA = \sqrt{3}$ , then cos A cos C – sin A sin C =
â€‹
(a) –1

(b) 0

(c) 1

(d) $\frac{\sqrt{3}}{2}$ â€‹

Answer:

$tanA = \sqrt{3} \Rightarrow \tan A = \tan 60 ° \Rightarrow A = 60 °$

ΔABC right angled at B, by using Angle Sum Property of Triangle
$∠ A + ∠ B + ∠ C = 180 ° \Rightarrow 60 ° + 90 ° + ∠ C = 180 ° \Rightarrow ∠ C = 180 ° - 150 ° \Rightarrow ∠ C = 30 °$

cos A cos C – sin A sin C
$= \cos 60 ° \cos 30 ° - \sin 60 ° \sin 30 ° = \frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$

Hence, the correct answer is option (b).

Page No 155:

Question 5:

If the angle of ΔABC are in the ratio 1 : 1 : 2 respectively (the largest angle being angle C), then the value of $\frac{secA}{cosecB} - \frac{tanA}{cotB}$ is
(a) 0

(b) $\frac{1}{2}$

(c) 1

(d) $\frac{\sqrt{3}}{2}$ â€‹

Answer:

The angle of ΔABC are in the ratio 1 : 1 : 2.

Thus, A = x, B = x and C = 2x.

In ΔABC, by using Angle Sum Property of Triangle
∠A + ∠B + ∠C = 180^âˆ˜
⇒ x + x + 2x = 180^âˆ˜
⇒ 4x = 180^âˆ˜
⇒ x = 45^âˆ˜

Thus, A = B = 45^âˆ˜ and C = 90^âˆ˜.
$\begin{array}{rcl} \frac{secA}{cosecB} - \frac{tanA}{cotB} & = & \frac{\sec 45 °}{cosec 45 °} - \frac{\tan 45 °}{\cot 45 °} \\ = & \frac{\sqrt{2}}{\sqrt{2}} - \frac{1}{1} \\ = & 0 \end{array}$

Hence, the correct answer is option (a).

Page No 156:

Question 6:

If θ is an acute angle such that $\cos θ = \frac{3}{5}, then \frac{\sin θ \tan θ - 1}{2 \tan^{2} θ} =$

(a) $\frac{16}{625}$
(b) $\frac{1}{36}$
(c) $\frac{3}{160}$
(d) $\frac{160}{3}$

Answer:

Given: and we need to find the value of the following expression

We know that:

So we find,

Hence the correct option is

Page No 156:

Question 7:

If $\tan θ = \frac{a}{b}, then \frac{a \sin θ + b \cos θ}{a \sin θ - b \cos θ}$ is equal to

(a) $\frac{a^{2} + b^{2}}{a^{2} - b^{2}}$
(b) $\frac{a^{2} - b^{2}}{a^{2} + b^{2}}$
(c) $\frac{a + b}{a - b}$
(s) $\frac{a - b}{a + b}$

Answer:

Given:

We have to find the value of following expression in terms of a and b

We know that:

Now we find,

Hence the correct option is

Page No 156:

Question 8:

If 5 tan θ − 4 = 0, then the value of $\frac{5 \sin θ - 4 \cos θ}{5 \sin θ + 4 \cos θ}$ is

(a) $\frac{5}{3}$
(b) $\frac{5}{6}$
(c) 0
(d) $\frac{1}{6}$

Answer:

Given that:.We have to find the value of the following expression

Since

We know that:

Since and

Now we find

Hence the correct option is

Page No 156:

Question 9:

If 16 cot x = 12, then $\frac{\sin x - \cos x}{\sin x + \cos x}$ equals

(a) $\frac{1}{7}$
(b) $\frac{3}{7}$
(c) $\frac{2}{7}$
(d) 0

Answer:

We are given .We are asked to find the following

We know that:

Now we have

We knowand

Now we find

Hence the correct option is

Page No 156:

Question 10:

If 8 tan x = 15, then sin x − cos x is equal to

(a) $\frac{8}{17}$
(b) $\frac{17}{7}$
(c) $\frac{1}{17}$
(d) $\frac{7}{17}$

Answer:

Given that:

We know that and

We find:

Hence the correct option is

Page No 156:

Question 11:

If $\tan θ = \frac{1}{\sqrt{7}}, then \frac{{cosec}^{2} θ - \sec^{2} θ}{{cosec}^{2} θ + \sec^{2} θ} =$

(a) $\frac{5}{7}$
(b) $\frac{3}{7}$
(c) $\frac{1}{12}$
(d) $\frac{3}{4}$

Answer:

Given that:

We are asked to find the value of the following expression

Since

We know that and

We find:

Hence the correct option is

Page No 156:

Question 12:

If $\tan θ = \frac{3}{4}$ , then cos² θ − sin² θ =

(a) $\frac{7}{25}$
(b) 1
(c) $\frac{- 7}{25}$
(d) $\frac{4}{25}$

Answer:

Given that:

Since

We know that and

We find:

Hence the correct option is

Page No 156:

Question 13:

If θ is an acute angle such that $\tan^{2} θ = \frac{8}{7}$ , then the value of $\frac{(1 + \sin θ) (1 - \sin θ)}{(1 + \cos θ) (1 - \cos θ)}$ is

(a) $\frac{7}{8}$
(b) $\frac{8}{7}$
(c) $\frac{7}{4}$
(d) $\frac{64}{49}$

Answer:

Given that: andis an acute angle

We have to find the following expression

Since

We know thatand

We find:

Hence the correct option is

Page No 156:

Question 14:

If 3 cos θ = 5 sin θ, then the value of $\frac{5 \sin θ - 2 \sec^{3} θ + 2 \cos θ}{5 \sin θ + 2 \sec^{3} θ - 2 \cos θ}$ is

(a) $\frac{271}{979}$
(b) $\frac{316}{2937}$
(c) $\frac{542}{2937}$
(d) None of these

Answer:

We have,

So we can manipulate it as,

So now we can get the values of other trigonometric ratios,

So now we will put these values in the equation,

So the answer is (a).

Page No 156:

Question 15:

If tan² 45° − cos² 30° = x sin 45° cos 45°, then x =

(a) 2
(b) −2
(c) $- \frac{1}{2}$
(d) $\frac{1}{2}$

Answer:

We are given:

We have to find x

We know that

Hence the correct option is

Page No 156:

Question 16:

The value of cos² 17° − sin² 73° is

(a) 1
(b) $\frac{1}{3}$
(c) 0
(d) −1

Answer:

We have:

Hence the correct option is

Page No 156:

Question 17:

The value of $\frac{\cos^{3} 20 ° - \cos^{3} 70 °}{\sin^{3} 70 ° - \sin^{3} 20 °}$ is

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

Answer:

We have to evaluate the value. The formula to be used,

So,

Now using the properties of complementary angles,

So the answer is

Page No 157:

Question 18:

If $\frac{x {cosec}^{2} 30 ° \sec^{2} 45 °}{8 \cos^{2} 45 ° \sin^{2} 60 °} = \tan^{2} 60 ° - \tan^{2} 30 °$ , then x =

(a) 1
(b) −1
(c) 2
(d) 0

Answer:

We have:

Here we have to find the value of

As we know that

Hence the correct option is

Page No 157:

Question 19:

If A and B are complementary angles, then

(a) sin A = sin B
(b) cos A = cos B
(c) tan A = tan B
(d) sec A = cosec B

Answer:

Given: and are are complementary angles

Since

Hence the correct option is

Page No 157:

Question 20:

If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =

(a) 0
(b) 1
(c) −1
(d) 2

Answer:

We have:

Here we have to find the value of

We know that

Hence the correct option is

Page No 157:

Question 21:

If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to

(a) 1
(b) $\sqrt{3}$
(c) $\frac{1}{2}$
(d) $\frac{1}{\sqrt{2}}$

Answer:

Given that:

Here we have to find the value of

We know that

Hence the correct option is

Page No 157:

Question 22:

If angles A, B, C to a âˆ†ABC from an increasing AP, then sin B =

(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) 1
(d) $\frac{1}{\sqrt{2}}$

Answer:

Let the angles of a triangleberespectively which constitute an A.P.As we know that sum of all the three angles of a triangle is. So,

So,

Therefore,

Hence,

So answer is

Page No 157:

Question 23:

If θ is an acute angle such that sec² θ = 3, then the value of $\frac{\tan^{2} θ - {cosec}^{2} θ}{\tan^{2} θ + {cosec}^{2} θ}$ is

(a) $\frac{4}{7}$
(b) $\frac{3}{7}$
(c) $\frac{2}{7}$
(d) $\frac{1}{7}$

Answer:

Given that:

We need to find the value of the expression

.So

Here we have to find:

Hence the correct option is

Page No 157:

Question 24:

The value of tan 1° tan 2° tan 3° ...... tan 89° is

(a) 1
(b) −1
(c) 0
(d) None of these

Answer:

Here we have to find:

Hence the correct option is

Page No 157:

Question 25:

The value of cos 1° cos 2° cos 3° ..... cos 180° is

(a) 1
(b) 0
(c) −1
(d) None of these

Answer:

Here we have to find:

Hence the correct option is

Page No 157:

Question 26:

The value of tan 10° tan 15° tan 75° tan 80° is

(a) −1
(b) 0
(c) 1
(d) None of these

Answer:

Here we have to find:

Now

Hence the correct option is

Page No 157:

Question 27:

The value of $\frac{\cos (90 ° - θ) \sec (90 ° - θ) \tan θ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{\cot θ}$ is

(a) 1
(b) − 1
(c) 2
(d) −2

Answer:

We have to find:

Hence the correct option is

Page No 157:

Question 28:

If θ and 2θ − 45° are acute angles such that sin θ = cos (2θ − 45°), then tan θ is equal to

(a) 1
(b) −1
(c) $\sqrt{3}$
(d) $\frac{1}{\sqrt{3}}$

Answer:

Given that: and are acute angles

We have to find

Where and are acute angles

Since

Now

Put

Hence the correct option is

Page No 157:

Question 29:

If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ − $\sqrt{3} \tan 3 θ$ is equal to

(a) 1
(b) 0
(c) −1
(d) $1 + \sqrt{3}$

Answer:

We are given that and are acute angles satisfying the following condition

.We are asked to find

Where and are acute angles

Now we have to find:

Hence the correct option is

Page No 157:

Question 30:

If A + B = 90°, then $\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^{2} B}{\cos^{2} A}$ is equal to

(a) cot² A
(b) cot² B
(c) −tan² A
(d) −cot² A

Answer:

We have:

We have to find the value of the following expression

Hence the correct option is

Page No 157:

Question 31:

$\frac{2 \tan 30 °}{1 + \tan^{2} 30 °}$ is equal to

(a) sin 60°
(b) cos 60°
(c) tan 60°
(d) sin 30°

Answer:

We have to find the value of the following expression

Hence the correct option is

Page No 157:

Question 32:

$\frac{1 - \tan^{2} 45 °}{1 + \tan^{2} 45 °}$ is equal to

(a) tan 90°
(b) 1
(c) sin 45°
(d) sin 0°

Answer:

We have to find the value of the following

We know that

Hence the correct option is

Page No 157:

Question 33:

Sin 2A = 2 sin A is true when A =

(a) 0°
(b) 30°
(c) 45°
(d) 60°

Answer:

We are given

Hence the correct option is

Page No 158:

Question 34:

$\frac{2 \tan 30 °}{1 - \tan^{2} 30 °}$ is equal to

(a) cos 60°
(b) sin 60°
(c) tan 60°
(d) sin 30°

Answer:

We are asked to find the value of the following

We know that

Hence the correct option is

Page No 158:

Question 35:

If A, B and C are interior angles of a triangle ABC, then $\sin (\frac{B + C}{2}) =$

(a) $\sin \frac{A}{2}$
(b) $\cos \frac{A}{2}$
(c) $- \sin \frac{A}{2}$
(d) $- \cos \frac{A}{2}$

Answer:

We know that in triangle

Hence the correct option is

Page No 158:

Question 36:

If $\cos θ = \frac{2}{3}$ , then 2 sec² θ + 2 tan² θ − 7 is equal to

(a) 1
(b) 0
(c) 3
(d) 4

Answer:

Given that:

We have to find

As we are given

We know that:

Now we have to find: .So

Hence the correct option is

Page No 158:

Question 37:

tan 5° âœ• tan 30° âœ• 4 tan 85° is equal to

(a) $\frac{4}{\sqrt{3}}$
(b) $4 \sqrt{3}$
(c) 1
(d) 4

Answer:

We have to find

We know that

Hence the correct option is

Page No 158:

Question 38:

The value of $\frac{\tan 55 °}{\cot 35 °}$ + cot 1° cot 2° cot 3° .... cot 90°, is

(a) −2
(b) 2
(c) 1
(d) 0

Answer:

We have to find the value of the following expression

Hence the correct option is

Page No 158:

Question 39:

In Fig. 5.47, the value of cos Ï• is

(a) $\frac{5}{4}$
(b) $\frac{5}{3}$
(c) $\frac{3}{5}$
(d) $\frac{4}{5}$

Answer:

We should proceed with the fact that sum of angles on one side of a straight line is.

So from the given figure,

So, …… (1)

Now from the triangle,

Now we will use equation (1) in the above,

Therefore,

So the answer is

Page No 158:

Question 40:

In Fig. 5.48, AD = 4 cm, BD = 3 cm and CB = 12 cm, find the cot θ.

(a) $\frac{12}{5}$
(b) $\frac{5}{12}$
(c) $\frac{13}{12}$
(d) $\frac{12}{13}$

Answer:

We have the following given data in the figure,

Now we will use Pythagoras theorem in,

Therefore,

So the answer is

Page No 158:

Question 41:

In the given figure, if D is the mid-point of BC, then the value of $\frac{\cot y °}{\cot x °}$ is

(a) 2

(b) $\frac{1}{2}$

(c) $\frac{1}{3}$

â€‹(d) $\frac{3}{4}$

Answer:

In âˆ†ACB,
$\begin{array}{rcl} \cot y ° & = & \frac{AC}{BC} \\ = & \frac{AC}{2 CD} . . . . . (1) \end{array}$

In âˆ†ACD,
$\cot x ° = \frac{AC}{CD} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} . \end{array} \begin{array}{rcl} (2) \end{array}$

Dividing (1) by (2), we get

$∴ \frac{\cot y °}{\cot x °} = \frac{AC}{2 CD} \times \frac{CD}{AC} = \frac{1}{2} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array} \begin{array}{rcl} \end{array}$

Hence, the correct answer is option (b).

Page No 158:

Question 42:

In structural design a structure is composed of triangles that are interconnecting. A truss is one of the major types of engineering structures and is especially used in the design of bridges and buildings. Trusses are designed to support loads, such as the weight of people. A truss is exclusively made of long, straight members connected by joints at the end of each member.

This is a single repeating triangle in a truss system.

(i) In above triangle, what is the length of AC?

(a) 5 ft

(b) 6 ft

(c) 8 ft

(d)

\frac{8}{\sqrt{3}} ft

(ii) What is the length of BC?

(a)

\frac{4}{\sqrt{3}} ft

(b)

4 \sqrt{3} ft

3 \sqrt{3} ft

(iii) If sinA = sinC, what will be the length of BC?

(a) 2ft

(b) 4 ft

(c) 8 ft

(d)

4 \sqrt{2} ft

(iv) Which of the following relation will be true in the triangle?

(a)

\sin (\frac{A + C}{2}) = \cos (\frac{B}{2})

(b)

\sin (\frac{A + B}{2}) = \sin (\frac{C}{2})

(c)

\cos (\frac{A + B}{2}) = \cos (\frac{C}{2})

(d)

\cos (\frac{A - B}{2}) = \cos (\frac{C}{2})

(v) If the length of AB doubles what will happen to the length of Aâ€‹C?

â€‹(a) remains same

(b) doubles the original length

(c) become three times the original length

(d) become half of the original length

Answer:

(i) In âˆ†ABC,
$\sin 30 ° = \frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{4}{AC} \Rightarrow AC = 8 ft$

Hence, the correct answer is option (c).

(ii) In âˆ†ABC,
$\tan 30 ° = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{4}{BC} \Rightarrow BC = 4 \sqrt{3} ft$

Hence, the correct answer is option (b).

(iii) sinA = sinC
$\Rightarrow \frac{BC}{AC} = \frac{AB}{AC} \Rightarrow BC = AB = 4 ft$

Hence, the correct answer is option (b).

(iv) In âˆ†ABC,
A + B + C = 180^âˆ˜ (âˆµ Angle Sum Property )
$\Rightarrow \frac{A + B + C}{2} = 90 ° \Rightarrow \frac{A + C}{2} = 90 ° - \frac{B}{2} \Rightarrow \sin (\frac{A + C}{2}) = \sin (90 ° - \frac{B}{2}) [∵ \sin (90 ° - θ) = \cos (θ)] \Rightarrow \sin (\frac{A + C}{2}) = \cos (\frac{B}{2})$
Hence, the correct answer is option (a).

(v) If length of AB is double, then AB = 8 ft
In âˆ†ABC,
$\sin 30 ° = \frac{AB}{AC} \Rightarrow \frac{1}{2} = \frac{8}{AC} \Rightarrow AC = 16 ft$
Thus, the length of AC is also doubles the original length.

Hence, the correct answer is option (b).

Page No 159:

Question 43:

A trolley carries passengers from the ground level located at point A to up to the top of mountain chateau located at P as shown in the given figure. The point A is at a distance of 2000 m from point C at the base of mountain. Here α = 30°, β = 60°.

(i) Assuming the cable is held tight what will be the length of cable?

(a) 2000 m

(b)

2000 \sqrt{3} m

(c)

4000 \sqrt{3} m

(d)

\frac{4000}{\sqrt{3}} m

(ii) What will be height of the mountain?

(a) 1000 m

(b)

\frac{2000}{\sqrt{3}} m

2000 \sqrt{3} m

(iii) What will be the slant height of the mountain?

(a) 4000 m

(b)

\frac{4000}{3} m

(c)

4000 \sqrt{3} m

(d)

\frac{4000}{\sqrt{3}} m

(iv) What will be the length of BC?

(a) 1000 m

(b)

\frac{2000}{3} m

(c)

1000 \sqrt{3} m

(d)

\frac{1000}{\sqrt{3}} m

(v) What will be the distance of point A to the foot of the mountain located at B?

â€‹(a)

4000 \sqrt{3} m

(b)

4000 \sqrt{6} m

(c)

\frac{4000}{\sqrt{3}} m

(d)

\frac{4000}{3} m

â€‹

Answer:

(i) In âˆ†PAC,
$cosα = \frac{AC}{AP} \Rightarrow \cos 30 ° = \frac{2000}{AP} \Rightarrow \frac{\sqrt{3}}{2} = \frac{2000}{AP} \Rightarrow AP = \frac{4000}{\sqrt{3}} m$

Hence, the correct answer is option (d).

(ii) In âˆ†PAC,
$tanα = \frac{PC}{AC} \Rightarrow \tan 30 ° = \frac{PC}{2000} \Rightarrow \frac{1}{\sqrt{3}} = \frac{PC}{2000} \Rightarrow PC = \frac{2000}{\sqrt{3}} m$

Hence, the correct answer is option (b).

(iii) In âˆ†PBC,
$sinβ = \frac{PC}{PB} \Rightarrow \sin 60 ° = \frac{2000}{\sqrt{3} PB} [From (ii)] \Rightarrow \frac{\sqrt{3}}{2} = \frac{2000}{PB \sqrt{3}} \Rightarrow PB = \frac{4000}{3} m$

Hence, the correct answer is option (b).

(iv) In âˆ†PBC,
$tanβ = \frac{PC}{BC} \Rightarrow \tan 60 ° = \frac{2000}{\sqrt{3} BC} [From (ii)] \Rightarrow \sqrt{3} = \frac{2000}{\sqrt{3} BC} \Rightarrow BC = \frac{2000}{3} m$

Hence, the correct answer is option (b).

(v)
$\begin{array}{rcl} AB & = & AC - BC \\ = & 2000 - \frac{2000}{3} [From (ii)] \\ = & \frac{4000}{3} m \end{array}$

Hence, the correct answer is option (d).

Page No 160:

Question 44:

â€‹â€‹â€‹â€‹â€‹Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): For any acute angle θ, values of tan θ never exceeds

\sqrt{3}

.
Statement-2 (Reason): For

0 \leq θ < 90 °, tanθ = \frac{sinθ}{cosθ} .

Answer:

Statement-2 (Reason): For $0 \leq θ < 90 °, tanθ = \frac{sinθ}{cosθ} .$
For $0 \leq θ < 90 °, tanθ = \frac{sinθ}{cosθ} .$
Thus, statement-2 is true.

Statement-1 (Assertion): For any acute angle θ, values of tan θ never exceeds $\sqrt{3}$ .
For any θ, tan θ ∈ (0, ∞).
Thus, statement-1 is false.

Hence, the correct answer is option (d).

Page No 160:

Question 45:

Answer:

Statement-2 (Reason): For any acute angle θ(0 < θ ≤ 90°), cosec θâ€‹ ≥ 1
For any acute angle θ(0 < θ ≤ 90°),
0 ≤ sin θ ≤ 1
⇒ cosec θâ€‹ ≥ 1
Thus, Statement-2 is true.

Statement-1 (Assertion): For any acute angle θ(0 ≤ θ < 90°), sec θ ≥ 1
For any acute angle θ(0 ≤ θ < 90°),
0 ≤ cos θ ≤ 1
⇒ sec θ ≥ 1
Thus, Statement-1 is true.
But Statement-2 is not a correct explanation for Statement-1.

Thus, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Page No 160:

Question 46:

â€‹â€‹â€‹â€‹â€‹â€‹Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

x + \frac{1}{x} \geq 2 for all x > 0

Answer:

Statement-2 (Reason): $x + \frac{1}{x} \geq 2 for all x > 0$ .
For any x > 0.
Apply AM ≥ GM on $x and \frac{1}{x}$ ,

$\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \times \frac{1}{x}} \Rightarrow x + \frac{1}{x} \geq 2$

Thus, statement-2 is true.

Statement-1 (Assertion): For 0 < θ ≤ 90°, sin θ + cosec θ ≥ 2.
In Statement-1 putting x = sin θ
$\Rightarrow \sin θ + \frac{1}{\sin θ} \geq 2 \Rightarrow \sin θ + \cos e c θ \geq 2$

Thus, Statement-1 is true and Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

View NCERT Solutions for all chapters of Class 10