Rs Aggarwal 2019 2020 for Class 9 Math Chapter 8 - Triangles

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Rs Aggarwal 2019 2020 Solutions for Class 9 Math Chapter 8 Triangles are provided here with simple step-by-step explanations. These solutions for Triangles are extremely popular among Class 9 students for Math Triangles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 9 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

Page No 252:

Question 1:

In ∆ABC, if ∠B = 76° and ∠C = 48°, find ∠A.

Answer:

$In ∆ A B C, ∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow ∠ A + 76 ° + 48 ° = 180 ° \Rightarrow ∠ A + 124 ° = 180 ° \Rightarrow ∠ A = 56 °$

Page No 252:

Question 2:

The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.

Answer:

Let the angles of the given triangle measure $(2 x) °, (3 x) ° and (4 x) °$ , respectively.
Then,
$2 x + 3 x + 4 x = 180 ° [Sum of the angles of a triangle] \Rightarrow 9 x = 180 ° \Rightarrow x = 20 °$
Hence, the measures of the angles are $2 \times 20 ° = 40 °, 3 \times 20 ° = 60 ° and 4 \times 20 ° = 80 °$ .

Page No 252:

Question 3:

In ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate ∠A, ∠B and ∠C.

Answer:

Let $3 ∠ A = 4 ∠ B = 6 ∠ C = x °$ .
Then,
$∠ A = {(\frac{x}{3})}^{°}, ∠ B = {(\frac{x}{4})}^{°} and ∠ C = {(\frac{x}{6})}^{°} ∴ \frac{x}{3} + \frac{x}{4} + \frac{x}{6} = 180 ° [Sum of the angles of a triangle] \Rightarrow 4 x + 3 x + 2 x = 2160 ° \Rightarrow 9 x = 2160 ° \Rightarrow x = 240 °$
Therefore,
$∠ A = {(\frac{240}{3})}^{°} = 80 °, ∠ B = {(\frac{240}{4})}^{°} = 60 ° and ∠ C = {(\frac{240}{6})}^{°} = 40 °$

Page No 252:

Question 4:

In ∆ABC, if ∠A + ∠B = 108° and ∠B + ∠C = 130°, find ∠A, ∠B and ∠C.

Answer:

Let $∠ A + ∠ B = 108 ° and ∠ B + ∠ C = 130 °$ .
$\Rightarrow ∠ A + ∠ B + ∠ B + ∠ C = (108 + 130) ° \Rightarrow (∠ A + ∠ B + ∠ C) + ∠ B = 238 ° [∵ ∠ A + ∠ B + ∠ C = 180 °] \Rightarrow 180 ° + ∠ B = 238 ° \Rightarrow ∠ B = 58 °$

$∴ ∠ C = 130 ° - ∠ B$
$= (130 - 58) ° = 72 °$

$∴ ∠ A = 108 ° - ∠ B$
$= (108 - 58) ° = 50 °$

Page No 252:

Question 5:

In ∆ABC, ∠A + ∠B = 125° and ∠A + ∠C = 113°. Find ∠A, ∠B and ∠C.

Answer:

Let $∠ A + ∠ B = 125 ° and ∠ A + ∠ C = 113 °$ .
Then,
$∠ A + ∠ B + ∠ A + ∠ C = (125 + 113) ° \Rightarrow (∠ A + ∠ B + ∠ C) + ∠ A = 238 ° \Rightarrow 180 ° + ∠ A = 238 ° \Rightarrow ∠ A = 58 °$

$∴ ∠ B = 125 ° - ∠ A$
$= (125 - 58) ° = 67 °$

$∴ ∠ C = 113 ° - ∠ A$
$= (113 - 58) ° = 55 °$

Page No 252:

Question 6:

In ∆PQR, if ∠P − ∠Q = 42° and ∠Q − ∠R = 21°, find ∠P, ∠Q and ∠R.

Answer:

$Given : ∠ P - ∠ Q = 42 ° and ∠ Q - ∠ R = 21 °$
Then,
$∠ P = 42 ° + ∠ Q and ∠ R = ∠ Q - 21 ° ∴ 42 ° + ∠ Q + ∠ Q + ∠ Q - 21 ° = 180 ° [Sum of the angles of a triangle] \Rightarrow 3 ∠ Q = 159 ° \Rightarrow ∠ Q = 53 °$

$∴ ∠ P = 42 ° + ∠ Q$
$= (42 + 53) ° = 95 °$

$∴ ∠ R = ∠ Q - 21 °$
$= (53 - 21) ° = 32 °$

Page No 252:

Question 7:

The sum of two angles of a triangle is 116° and their difference is 24°. Find the measure of each angle of the triangle.

Answer:

Let $∠ A + ∠ B = 116 ° and ∠ A - ∠ B = 24 °$
Then,
$∴ ∠ A + ∠ B + ∠ A - ∠ B = (116 + 24) ° \Rightarrow 2 ∠ A = 140 ° \Rightarrow ∠ A = 70 °$

$∴ ∠ B = 116 ° - ∠ A$
$= (116 - 70) ° = 46 °$

Also, in ∆ ABC:
$∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow 70 ° + 46 ° + ∠ C = 180 ° \Rightarrow ∠ C = 64 °$

Page No 252:

Question 8:

Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.

Answer:

Let $∠ A = ∠ B and ∠ C = ∠ A + 18 °$ .
Then,
$∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] ∠ A + ∠ A + ∠ A + 18 ° = 180 ° \Rightarrow 3 ∠ A = 162 ° \Rightarrow ∠ A = 54 °$
Since,
$∠ A = ∠ B \Rightarrow ∠ B = 54 ° ∴ ∠ C = ∠ A + 18 °$

$= (54 + 18) ° = 72 °$

Page No 252:

Question 9:

Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Answer:

Let the smallest angle of the triangle be $∠ C$ and let $∠ A = 2 ∠ C and ∠ B = 3 ∠ C$ .
Then,
$∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow 2 ∠ C + 3 ∠ C + ∠ C = 180 ° \Rightarrow 6 ∠ = 180 ° \Rightarrow ∠ C = 30 °$

$∴ ∠ A = 2 ∠ C$
$= 2 (30) ° = 60 °$
Also,
$∠ B = 3 ∠ C = 3 (30) ° = 90 °$

Page No 252:

Question 10:

In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.

Answer:

Let ABC be a triangle right-angled at B.
Then, $∠ B = 90 ° and let ∠ A = 53 ° .$
$∴ ∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow 53 ° + 90 ° + ∠ C = 180 ° \Rightarrow ∠ C = 37 °$
Hence, $∠ A = 53 °, ∠ B = 90 ° and ∠ C = 37 °$ .

Page No 252:

Question 11:

If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.

Answer:

Let ABC be a triangle.
Then, $∠ A = ∠ B + ∠ C$
$∴ ∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow ∠ B + ∠ C + ∠ B + ∠ C = 180 ° \Rightarrow 2 ∠ B + ∠ C = 180 ° \Rightarrow ∠ B + ∠ C = 90 ° \Rightarrow ∠ A = 90 ° [∵ ∠ A = ∠ B + ∠ C]$
This implies that the triangle is right-angled at A.

Page No 252:

Question 12:

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.

Answer:

Let ABC be the triangle.
Let $∠ A < ∠ B + ∠ C$
Then,
$2 ∠ A < ∠ A + ∠ B + ∠ C [Adding ∠ A to both sides] \Rightarrow 2 ∠ A < 180 ° [∵ ∠ A + ∠ B + ∠ C = 180 °] \Rightarrow ∠ A < 90 °$

Also, let $∠ B < ∠ A + ∠ C$
Then,
$2 ∠ B < ∠ A + ∠ B + ∠ C [Adding ∠ B to both sides] \Rightarrow 2 ∠ B < 180 ° [∵ ∠ A + ∠ B + ∠ C = 180 °] \Rightarrow ∠ B < 90 °$

And let $∠ C < ∠ A + ∠ B$
Then,
$2 ∠ C < ∠ A + ∠ B + ∠ C [Adding ∠ C to both sides] \Rightarrow 2 ∠ C < 180 ° [∵ ∠ A + ∠ B + ∠ C = 180 °] \Rightarrow ∠ C < 90 °$

Hence, each angle of the triangle is less than $90 °$ .
Therefore, the triangle is acute-angled.

Page No 252:

Question 13:

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.

Answer:

Let ABC be a triangle and let $∠ C > ∠ A + ∠ B$ .
Then, we have:
$2 ∠ C > ∠ A + ∠ B + ∠ C [Adding ∠ C to both sides] \Rightarrow 2 ∠ C > 180 ° [∵ ∠ A + ∠ B + ∠ C = 180 °] \Rightarrow ∠ C > 90 °$
Since one of the angles of the triangle is greater than $90 °$ , the triangle is obtuse-angled.

Page No 252:

Question 14:

In the given figure, side BC of ∆ABC is produced to D. If ∠ACD = 128° and ∠ABC = 43°, find ∠BAC and ∠ACB.

Answer:

Side BC of triangle ABC is produced to D.
$∴ ∠ A C D = ∠ A + ∠ B [Exterior angle property] \Rightarrow 128 ° = ∠ A + 43 ° \Rightarrow ∠ A = (128 - 43) ° \Rightarrow ∠ A = 85 ° \Rightarrow ∠ B A C = 85 °$
Also, in triangle ABC,
$∠ B A C + ∠ A B C + ∠ A C B = 180 ° [Sum of the angles of a triangle] \Rightarrow 85 ° + 43 ° + ∠ A C B = 180 ° \Rightarrow 128 ° + ∠ A C B = 180 ° \Rightarrow ∠ A C B = 52 °$

Page No 252:

Question 15:

In the given figure, the side BC of ∆ ABC has been produced on both sides−on the left to D and on the right to E. If ∠ABD = 106° and ∠ACE = 118°, find the measure of each angle of the triangle.

Answer:

Side BC of triangle ABC is produced to D.

$∴ ∠ A B C = ∠ A + ∠ C \Rightarrow 106 ° = ∠ A + ∠ C . . . (i)$

Also, side BC of triangle ABC is produced to E.

$∠ A C E = ∠ A + ∠ B \Rightarrow 118 ° = ∠ A + ∠ B . . . (i i)$

$Adding (i) and (i i), we get : ∠ A + ∠ A + ∠ B + ∠ C = (106 + 118) °$
$\Rightarrow (∠ A + ∠ B + ∠ C) + ∠ A = 224 ° [∵ ∠ A + ∠ B + ∠ C = 180 °] \Rightarrow 180 ° + ∠ A = 224 ° \Rightarrow ∠ A = 44 °$

$∴ ∠ B = 118 ° - ∠ A [Using (i i)] \Rightarrow ∠ B = (118 - 44) ° \Rightarrow ∠ B = 74 °$

And,

$∠ C = 106 ° - ∠ A [Using (i)] \Rightarrow ∠ C = (106 - 44) ° \Rightarrow ∠ C = 62 °$

Page No 253:

Question 16:

Calculate the value of x in each of the following figures.

Answer:

(i)
Side AC of triangle ABC is produced to E.
$∴ ∠ E A B = ∠ B + ∠ C \Rightarrow 110 ° = x + ∠ C . . . (i)$
Also,
$∠ A C D + ∠ A C B = 180 ° [linear pair] \Rightarrow 120 ° + ∠ A C B = 180 ° \Rightarrow ∠ A C B = 60 ° \Rightarrow ∠ C = 60 °$
Substituting the value of $∠ C in (i), we get x = 50$

(ii)
From $∆ A B C$ we have:
$∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow 30 ° + 40 ° + ∠ C = 180 ° \Rightarrow ∠ C = 110 ° \Rightarrow ∠ A C B = 110 °$
Also,
$∠ E C B + ∠ E C D = 180 ° [linear pair] \Rightarrow 110 ° + ∠ E C D = 180 ° \Rightarrow ∠ E C D = 70 ° Now, in ∆ E C D, ∴ ∠ A E D = ∠ E C D + ∠ E D C [exterior angle property] \Rightarrow x = 70 ° + 50 ° \Rightarrow x = 120 °$

(iii)
$∠ A C B + ∠ A C D = 180 ° [linear pair] \Rightarrow ∠ A C B + 115 ° = 180 ° \Rightarrow ∠ A C B = 65 °$
Also,
$∠ E A F = ∠ B A C [Vertically-opposite angles] \Rightarrow ∠ B A C = 60 ° ∴ ∠ B A C + ∠ A B C + ∠ A C B = 180 ° [Sum of the angles of a triangle] \Rightarrow 60 ° + x + 65 ° = 180 ° \Rightarrow x = 55 °$

(iv)
$∠ B A E = ∠ C D E [Alternate angles] \Rightarrow ∠ C D E = 60 ° ∴ ∠ E C D + ∠ C D E + ∠ C E D = 180 ° [Sum of the angles of a triangle] \Rightarrow 45 ° + 60 ° + x = 180 ° \Rightarrow x = 75 °$

(v)
From $∆ A B C$ , we have:
$∠ B A C + ∠ A B C + ∠ A C B = 180 ° [Sum of the angles of a triangle] \Rightarrow 40 ° + ∠ A B C + 90 ° = 180 ° \Rightarrow ∠ A B C = 50 °$
Also, from $∆ E B D$ , we have:
$∠ B E D + ∠ E B D + ∠ B D E = 180 ° [Sum of the angles of a triangle] \Rightarrow 100 ° + 50 ° + x = 180 ° [∵ ∠ A B C = ∠ E B D] \Rightarrow x = 30 °$

(vi)
From $∆ A B E$ , we have:
$∠ B A E + ∠ A B E + ∠ A E B = 180 ° [Sum of the angles of a triangle] \Rightarrow 75 ° + 65 ° + ∠ A E B = 180 ° \Rightarrow ∠ A E B = 40 ° ∴ ∠ A E B = ∠ C E D [Vertically-opposite angles] \Rightarrow ∠ C E D = 40 °$

Also, From $∆ C D E$ , we have
$∠ E C D + ∠ C D E + ∠ C E D = 180 ° [Sum of the angles of a triangle] \Rightarrow 110 ° + x + 40 ° = 180 ° \Rightarrow x = 30 °$

Page No 253:

Question 17:

In the figure given alongside, AB || CD, EF || BC, ∠BAC = 60º and ∠DHF = 50º. Find ∠GCH and ∠AGH.

Answer:

In the given figure, AB || CD and AC is the transversal.

∴ ∠ACD = ∠BAC = 60º (Pair of alternate angles)

Or ∠GCH = 60º

Now, ∠GHC = ∠DHF = 50º (Vertically opposite angles)

In ∆GCH,

∠AGH = ∠GCH + ∠GHC (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ ∠AGH = 60º + 50º = 110º

Page No 253:

Question 18:

Calculate the value of x in the given figure.

Answer:

Join A and D to produce AD to E.
Then,
$∠ C A D + ∠ D A B = 55 ° and ∠ C D E + ∠ E D B = x °$
Side AD of triangle ACD is produced to E.
$∴ ∠ C D E = ∠ C A D + ∠ A C D . . . (i)$    (Exterior angle property)
Side AD of triangle ABD is produced to E.
$∴ ∠ E D B = ∠ D A B + ∠ A B D . . . (i i)$    (Exterior angle property)
$Adding (i) and (ii) we get, ∠ C D E + ∠ E D B = ∠ C A D + ∠ A C D + ∠ D A B + ∠ A B D$

   $\Rightarrow x ° = (∠ C A D + ∠ D A B) + 30 ° + 45 ° \Rightarrow x ° = 55 ° + 30 ° + 45 ° \Rightarrow x ° = 130 ° \Rightarrow x = 130$

Page No 254:

Question 19:

In the given figure, AD divides ∠BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.

Answer:

$∠ B A C + ∠ C A E = 180 ° [∵ B E is a straight line] \Rightarrow ∠ B A C + 108 ° = 180 ° \Rightarrow ∠ B A C = 72 °$

Now, divide $72 °$ in the ratio 1 : 3.
$∴ a + 3 a = 72 ° \Rightarrow a = 18 ° ∴ a = 18 ° and 3 a = 54 °$
Hence, the angles are 18^o and 54^o
$∴ ∠ B A D = 18 ° and ∠ D A C = 54 °$

Given,
$A D = D B \Rightarrow ∠ D A B = ∠ D B A = 18 °$

In $∆ A B C$ , we have:
$∠ B A C + ∠ A B C + ∠ A C B = 180 ° [Sum of the angles of a triangle] \Rightarrow 72 ° + 18 ° + x ° = 180 ° \Rightarrow x ° = 90 ° ∴ x = 90$

Page No 254:

Question 20:

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Answer:

Side BC of triangle ABC is produced to D.
$∠ A C D = ∠ B + ∠ A . . . (i)$
Side AC of triangle ABC is produced to E.
$∠ B A C = ∠ B + ∠ C . . . (i)$
And side AB of triangle ABC is produced to F.
$∠ C B F = ∠ C + ∠ A . . . (i i i)$

$Adding (i), (i i) and (i i i), we get: ∠ A C D + ∠ B A E + ∠ C B F = 2 (∠ A + ∠ B + ∠ C)$
$= 2 (180) ° = 360 ° = 4 \times 90 ° = 4 right angle s$
Hence, the sum of the exterior angles so formed is equal to four right angles.

Page No 254:

Question 21:

In the adjoining figure, show that
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°.

Answer:

In $∆ A C E$ , we have :
$∠ A + ∠ C + ∠ E = 180 ° . . . (i)$ [Sum of the angles of a triangle]
In $∆ B D F$ , we have :
$∠ B + ∠ D + ∠ F = 180 ° . . . (i i)$ [Sum of the angles of a triangle]

$Adding (i) and (i i), we get: ∠ A + ∠ C + ∠ E + ∠ B + ∠ D + ∠ F = (180 + 180) °$
$\Rightarrow ∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F = 360 °$

Page No 254:

Question 22:

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?

Answer:

In $∆ A B C$ , we have:
$∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow ∠ A + 70 ° + 20 ° = 180 ° \Rightarrow ∠ A = 90 ° \Rightarrow \frac{1}{2} ∠ A = 45 ° \Rightarrow ∠ B A N = 45 °$
In $∆ A B M$ , we have:
$∠ A B M + ∠ A M B + ∠ B A M = 180 ° [Sum of the angles of a triangle] \Rightarrow 70 ° + 90 ° + ∠ B A M = 180 ° \Rightarrow ∠ B A M = 20 ° ∴ ∠ M A N = ∠ B A N - ∠ B A M \Rightarrow ∠ M A N = 45 ° - 20 ° \Rightarrow ∠ M A N = 25 °$

Page No 254:

Question 23:

In the given figure, BAD || EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.

Answer:

In the given figure, EF || BD and CE is the transversal.

∴ ∠CAD = ∠AEF (Pair of corresponding angles)

⇒ ∠CAD = 55°

In ∆ABC,

∠CAD = ∠ABC + ∠ACB (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ 55° = ∠ABC + 25°

⇒ ∠ABC = 55° − 25° = 30°

Thus, the measure of ∠ABC is 30°.

Page No 255:

Question 24:

In the given figure, ABC is a triangle in which ∠A : ∠B : ∠C = 3 : 2 : 1 and AC ⊥ CD. Find the measure of ∠ECD.

Answer:

Let $∠ A = (3 x) °, ∠ B = (2 x) ° and ∠ C = x °$
From $∆ A B C$ , we have:
$∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow 3 x + 2 x + x = 180 ° \Rightarrow 6 x = 180 ° \Rightarrow x = 30 ° ∴ ∠ A = 3 (30) ° = 60 ° ∠ B = 2 (30) ° = 60 ° and ∠ C = 30 °$
Side BC of triangle ABC is produced to E.
$∴ ∠ A C E = ∠ A + ∠ B \Rightarrow ∠ A C D + ∠ E C D = (90 + 60) ° \Rightarrow 90 ° + ∠ E C D = 150 ° \Rightarrow ∠ E C D = 60 °$

Page No 255:

Question 25:

In the given figure, AB || DE and BD || FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y.

Answer:

We have, ∠FGE + ∠FGH = 180° (Linear pair of angles)

∴ y + 120° = 180°

⇒ y = 180° − 120° = 60°

Now, AB || DF and BD is the transversal.

∴ ∠ABD = ∠BDF (Pair of alternate angles)

⇒ ∠BDF = 50°

Also, BD || FG and DF is the transversal.

∴ ∠BDF = ∠DFG (Pair of alternate angles)

⇒ ∠DFG = 50° .....(1)

In ∆EFG,

∠FGH = ∠EFG + ∠FEG (Exterior angle of a triangle is equal to the sum of the two opposite interior angles)

⇒ 120° = 50° + x [Using (1)]

⇒ x = 120° − 50° = 70°

Thus, the values of x and y are 70° and 60°, respectively.

Page No 255:

Question 26:

In the given figure, AB || CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°, ∠EGF = 90° and ∠GEF = x°, find the value of x.

Answer:

It is given that, AB || CD and EF is a transversal.

∴ ∠EFD = ∠AEF (Pair of alternate angles)

⇒ ∠EFD = 65°

⇒ ∠EFG + ∠GFD = 65°

⇒ ∠EFG + 30° = 65°

⇒ ∠EFG = 65° − 30° = 35°

In ∆EFG,

∠EFG + ∠GEF + ∠EGF = 180° (Angle sum property)

⇒ 35° + x + 90° = 180°

⇒ 125° + x = 180°

⇒ x = 180° − 125° = 55°

Thus, the value of x is 55°.

Page No 255:

Question 27:

In the given figure, AB || CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.

Answer:

In the given figure, AB || CD and AE is the transversal.

∴ ∠DOE = ∠BAE (Pair of corresponding angles)

⇒ ∠DOE = 65°

In ∆COE,

∠DOE = ∠OEC + ∠ECO (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ 65° = 20° + ∠ECO

⇒ ∠ECO = 65° − 20° = 45°

Thus, the measure of ∠ECO is 45°.

Page No 255:

Question 28:

In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35° and QP ⊥ EF, find the measure of ∠PQH.

Answer:

In the given figure, AB || CD and EF is a transversal.

∴ ∠PHQ = ∠EGB (Pair of alternate exterior angles)

⇒ ∠PHQ = 35°

In ∆PHQ,

∠PHQ + ∠QPH + ∠PQH = 180° (Angle sum property)

⇒ 35° + 90° + x = 180°

⇒ 125° + x = 180°

⇒ x = 180° − 125° = 55°

Thus, the measure of ∠PQH is 55°.

Page No 255:

Question 29:

In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130°, find ∠EGF.

Answer:

In the given figure, AB || CD and GE is the transversal.

∴ ∠GED + ∠EGF = 180° (Sum of adjacent interior angles on the same side of the transversal is supplementary)

⇒ 130° + ∠EGF = 180°

⇒ ∠EGF = 180° − 130° = 50°

Thus, the measure of ∠EGF is 50°.

Page No 257:

Question 1:

In ∆ABC, if 3∠A = 4∠B = 6∠C then A : B : C = ?
(a) 3 : 4 : 6
(b) 4 : 3 : 2
(c) 2 : 3 : 4
(d) 6 : 4 : 3

Answer:

LCM of 3, 4 and 6 = 12

3∠A = 4∠B = 6∠C (Given)

Dividing throughout by 12, we get

$\frac{3 ∠ A}{12} = \frac{4 ∠ B}{12} = \frac{6 ∠ C}{12}$

$\Rightarrow \frac{∠ A}{4} = \frac{∠ B}{3} = \frac{∠ C}{2}$

Let $\frac{∠ A}{4} = \frac{∠ B}{3} = \frac{∠ C}{2} = k$ , where k is some constant

Then, ∠A = 4k, ∠B = 3k, ∠C = 2k

∴ ∠A : ∠B : ∠C = 4k : 3k : 2k = 4 : 3 : 2

Hence, the correct answer is option (b).

Page No 258:

Question 2:

In a ∆ABC, if ∠A − ∠B = 42° and ∠B − ∠C = 21° then ∠B = ?
(a) 32°
(b) 63°
(c) 53°
(d) 95°
Figure

Answer:

(c) 53°

Let
$∠ A - ∠ B = 42 ° . . . (i) and ∠ B - ∠ C = 21 ° . . . (i i) Adding (i) and (i i), we get: ∠ A - ∠ C = 63 ° ∠ B = ∠ A - 42 ° [Using (i)] ∠ C = ∠ A - 63 ° [Using (i i i)]$

$∴ ∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow ∠ A + ∠ A - 42 ° + ∠ A - 63 ° = 180 ° \Rightarrow 3 ∠ A - 105 ° = 180 ° \Rightarrow 3 ∠ A = 285 ° \Rightarrow ∠ A = 95 ° ∴ ∠ B = (95 - 42) ° \Rightarrow ∠ B = 53 °$

Page No 258:

Question 3:

In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ?
(a) 160°
(b) 60°
(c) 80°
(d) 30°

Answer:

$∴ ∠ A + ∠ B = ∠ A C D \Rightarrow ∠ A + 50 ° = 110 ° \Rightarrow ∠ A = 60 °$
Hence, the correct answer is option (b).

Page No 258:

Question 4:

Side BC of ∆ABC has been produced to D on left and to E on right hand side of BC such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?
(a) 50°
(b) 55°
(c) 65°
(d) 75°

Answer:

(d) 75°

We have :
$∴ ∠ A B D + ∠ A B C = 180 ° [∵ D E is a straight line] \Rightarrow 125 ° + ∠ A B C = 180 ° \Rightarrow ∠ A B C = 55 °$
Also,
$∠ A C E + ∠ A C B = 180 ° \Rightarrow 130 ° + ∠ A C B = 180 ° \Rightarrow ∠ A C B = 50 ° ∴ ∠ B A C + ∠ A B C + ∠ A C B = 180 ° [Sum of the angles of a triangle] \Rightarrow ∠ B A C + 55 ° + 50 ° = 180 ° \Rightarrow ∠ B A C = 75 ° \Rightarrow ∠ A = 75 °$

Page No 258:

Question 5:

In the given figure, the sides CB and BA of ∆ABC have been produced to D and E, respectively, such that ∠ABD = 110° and ∠CAE = 135°. Then ∠ACB = ?
(a) 65°
(b) 45°
(c) 55°
(d) 35°

Answer:

(a) 65°

We have :
$∠ A B D + ∠ A B C = 180 ° [∵ C B D is a straight line] \Rightarrow 100 ° + ∠ A B C = 180 ° \Rightarrow ∠ A B C = 70 °$
Side AB of triangle ABC is produced to E.
$∴ ∠ C A E = ∠ A B C + ∠ A C B \Rightarrow 135 ° = 70 ° + ∠ A C B \Rightarrow ∠ A C B = 65 °$

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Question 6:

The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively. ∠BAE + ∠CBF + ∠ACD = ?
(a) 240°
(b) 300°
(c) 320°
(d) 360°

Answer:

(d) 360°

We have :
$∠ 1 + ∠ B A E = 180 ° . . . (i) ∠ 2 + ∠ C B F = 180 ° . . . (i i) and ∠ 3 + ∠ A C D = 180 ° . . . (i i i)$
$Adding (i), (i i) and (i i i), we get: (∠ 1 + ∠ 2 + ∠ 3) + (∠ B A E + ∠ C B F + ∠ A C D) = 540 °$
$\Rightarrow 180 ° + ∠ B A E + ∠ C B F + ∠ A C D = 540 ° [∵ ∠ 1 + ∠ 2 + ∠ 3 = 180 °] \Rightarrow ∠ B A E + ∠ C B F + ∠ A C D = 360 °$

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Question 7:

The the given figure, EAD ⊥ BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF = 30°. Also, in ∆BAC, ∠BAC = x° and ∠ABC = (x + 10)°. Then, the value of x is
(a) 20
(b) 25
(c) 30
(d) 35

Answer:

In the given figure, ∠CAD = ∠EAF (Vertically opposite angles)

∴ ∠CAD = 30°

In ∆ABD,

∠ABD + ∠BAD + ∠ADB = 180° (Angle sum property)

⇒ (x + 10)° + (x° + 30°) + 90° = 180°

⇒ 2x° + 130° = 180°

⇒ 2x° = 180° − 130° = 50°

⇒ x = 25

Thus, the value of x is 25.

Hence, the correct answer is option (b).

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Question 8:

In the given figure, two rays BD and CE intersect at a point A. The side BC of ∆ABC have been produced on both sides to points F and G respectively. If ∠ABF = x°, ∠ACG = y° and ∠DAE = z° then z = ?
(a) x + y – 180
(b) x + y + 180
(c) 180 – (x + y)
(d) x + y + 360°

Answer:

In the given figure, ∠ABF + ∠ABC = 180° (Linear pair of angles)

∴ x° + ∠ABC = 180°

⇒ ∠ABC = 180° − x° .....(1)

Also, ∠ACG + ∠ACB = 180° (Linear pair of angles)

∴ y° + ∠ACB = 180°

⇒ ∠ACB = 180° − y° .....(2)

Also, ∠BAC = ∠DAE = z° .....(3) (Vertically opposite angles)

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property)

∴ z° + 180° − x° + 180° − y° = 180° [Using (1), (2) and (3)]

⇒ z = x + y − 180

Hence, the correct answer is option (a).

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Question 9:

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ∠OAE = x° and ∠DBF = y°.

If ∠OCA = 80°, ∠COA = 40° and ∠BDO = 70° then x° + y° = ?
(a) 190°
(b) 230°
(c) 210°
(d) 270°

Answer:

In the given figure, ∠BOD = ∠COA (Vertically opposite angles)

∴ ∠BOD = 40° .....(1)

In ∆ACO,

∠OAE = ∠OCA + ∠COA (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ x° = 80° + 40° = 120° .....(2)

In ∆BDO,

∠DBF = ∠BDO + ∠BOD (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ y° = 70° + 40° = 110° [Using (1)] .....(3)

Adding (2) and (3), we get

x° + y° = 120° + 110° = 230°

Hence, the correct answer is option (b).

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Question 10:

In a ∆ABC, it is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and ∠ACD = 90°. If BC is produced to E then ∠ECD = ?
(a) 60°
(b) 50°
(c) 40°
(d) 25°

Answer:

Let $∠ A = (3 x) °, ∠ B = (2 x) ° and ∠ C = x °$
Then,
$3 x + 2 x + x = 180 ° [Sum of the angles of a triangle] \Rightarrow 6 x = 180 ° \Rightarrow x = 30 °$
Hence, the angles are
$∠ A = 3 \times 30 ° = 90 °, ∠ B = 2 \times 30 ° = 60 ° and ∠ C = 30 °$
Side BC of triangle ABC is produced to E.
$∴ ∠ A C E = ∠ A + ∠ B \Rightarrow ∠ A C D + ∠ E C D = 90 ° + 60 ° \Rightarrow 90 ° + ∠ E C D = 150 ° \Rightarrow ∠ E C D = 60 °$
Hence, the correct answer is option (a).

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Question 11:

In the given figure, BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50° then ∠BOC = ?
(a) 130°
(b) 100°
(c) 115°
(d) 120°

Answer:

(c) 115°

In $∆ A B C$ , we have:
$∠ A + ∠ B + ∠ C = 180 ° [Sum of the angles of a triangle] \Rightarrow 50 ° + ∠ B + ∠ C = 180 ° \Rightarrow ∠ B + ∠ C = 130 ° \Rightarrow \frac{1}{2} ∠ B + \frac{1}{2} ∠ C = 65 ° . . . (i)$
In $∆ O B C$ , we have:
$∠ O B C + ∠ O C B + ∠ B O C = 180 ° \Rightarrow \frac{1}{2} ∠ B + \frac{1}{2} ∠ C + ∠ B O C = 180 ° [Using (i)] \Rightarrow 65 ° + ∠ B O C = 180 ° \Rightarrow ∠ B O C = 115 °$

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Question 12:

In the given figure, side BC of ∆ABC has been produced to a point D. If ∠A = 3y°, ∠B = x°, ∠C = 5y° and ∠CBD = 7y°. Then, the value of x is

(a) 60
(b) 50
(c) 45
(d) 35

Answer:

Disclaimer: In the question ∠ACD should be 7y°.

In the given figure, ∠ACB + ∠ACD = 180° (Linear pair of angles)

∴ 5y° + 7y° = 180°

⇒ 12y° = 180°

⇒ y = 15 .....(1)

In ∆ABC,

∠A + ∠B + ∠ACB = 180° (Angle sum property)

∴ 3y° + x° + 5y° = 180°

⇒ x° + 8y° = 180°

⇒ x° + 8 × 15° = 180° [Using (1)]

⇒ x° + 120° = 180°

⇒ x° = 180° − 120° = 60°

Thus, the value of x is 60.

Hence, the correct answer is option (a).

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