RD Sharma Solutions for Class 9 Maths Chapter 9 Triangle and its Angles

RD Sharma Class 9 Solutions is provided here for students to clear all their doubts regarding triangles as well as problems based on this topic. RD Sharma Solutions Class 9 Chapter 9 – Triangle and its Angles, includes three exercises and all questions are solved by BYJU’S subject experts. In this chapter, students will study about a plane figure formed by three non-parallel lines in a plane, called triangle. A triangle is a 2D geometrical figure consisting of three edges and three vertices. The three angles made by the three sides of a triangle and the sum of all the internal angles of a triangle is equal to 180 degrees. Triangles are categorized based on their angles and sides.

Triangles categorized based on their angles: Acute Triangle, Right Angle and Obtuse Triangle.

Triangles categorized based on their sides: Scalene triangle, Isosceles triangle and Equilateral triangle.

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Exercise 9.1 Page No: 9.9

Question 1: In a ΔABC, if ∠A = 55⁰, ∠B = 40⁰, find ∠C.

Solution:

Given: ∠A = 55⁰, ∠B = 40⁰

We know, sum of all angles of a triangle is 180⁰

∠A + ∠B + ∠C = 180⁰

55⁰ + 40⁰ + ∠C=180⁰

95⁰ + ∠C = 180⁰

∠C = 180⁰ − 95⁰

∠C = 85⁰

Question 2: If the angles of a triangle are in the ratio 1:2:3, determine three angles.

Solution:

Angles of a triangle are in the ratio 1:2:3 (Given)

Let the angles be x, 2x, 3x

Sum of all angles of triangles = 180⁰

x + 2x + 3x = 180⁰

6x = 180⁰

x = 180⁰/6

x = 30⁰

Answer:

x = 30⁰

2x = 2(30)⁰ = 60⁰

3x = 3(30) ⁰ = 90⁰

Question 3: The angles of a triangle are (x − 40)⁰, (x − 20) ⁰ and (1/2 x − 10) ⁰. Find the value of x.

Solution:

The angles of a triangle are (x − 40)⁰, (x − 20) ⁰ and (1/2 x − 10) ⁰

Sum of all angles of triangle = 180⁰

(x − 40)⁰ + (x − 20) ⁰ + (1/2 x − 10) ⁰ = 180⁰

5/2 x – 70⁰ = 1800

5/2 x = 180⁰ + 70⁰

5x = 2(250) ⁰

x = 500⁰/5

x = 100⁰

Question 4: The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10⁰, find the three angles.

Solution:

The difference between two consecutive angles is 10⁰ (given)

Let x, x + 10⁰, x + 20⁰ be the consecutive angles

x + x + 10⁰ + x + 20⁰ = 180⁰

3x + 30⁰ = 180⁰

3x = 180⁰– 30⁰

3x = 150⁰

or x = 50⁰

Again,

x + 10⁰ = 50⁰ + 10⁰ = 60⁰

x+20⁰ = 50⁰ + 20⁰ = 70⁰

Answer: Three angles are 50⁰,60⁰ and 70⁰.

Question 5: Two angles of a triangle are equal and the third angle is greater than each of those angles by 30⁰. Determine all the angles of the triangle.

Solution:

Two angles of a triangle are equal and the third angle is greater than each of those angles by 30⁰. (Given)

Let x, x, x + 30⁰ be the angles of a triangle.

Sum of all angles in a triangle = 180⁰

x + x + x + 30⁰ = 180⁰

3x + 30⁰ = 180⁰

3x = 150⁰

or x = 50⁰

And x + 30⁰ = 50⁰ + 30⁰ = 80⁰

Answer: Three angles are 50⁰, 50⁰ and 80⁰.

Question 6: If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right angle triangle.

Solution:

One angle of a triangle is equal to the sum of the other two angles (given)

To Prove: One of the angles is 90⁰

Let x, y and z are three angles of a triangle, where

z = x + y …(1)

Sum of all angles of a triangle = 180⁰

x + y + z = 180⁰

z + z = 180⁰ (Using equation (1))

2z = 180⁰

z = 90⁰ (Proved)

Therefore, triangle is a right angled triangle.

Exercise 9.2 Page No: 9.18

Question 1: The exterior angles, obtained on producing the base of a triangle both ways are 104⁰ and 136⁰. Find all the angles of the triangle.

Solution:

∠ACD = ∠ABC + ∠BAC [Exterior angle property]

Find ∠ABC:

∠ABC + ∠ABE = 180⁰ [Linear pair]

∠ABC + 136⁰ = 180⁰

∠ABC = 44⁰

Find ∠ACB:

∠ACB + ∠ACD = 180⁰ [Linear pair]

∠ACB + 104⁰ = 180⁰

∠ACB = 76⁰

Now,

Sum of all angles of a triangle = 180⁰

∠A + 44⁰ + 76⁰ = 180⁰

∠A = 180⁰ − 44⁰ −76⁰

∠ A = 60⁰

Answer: Angles of a triangle are ∠ A = 60⁰, ∠B = 44⁰ and ∠C = 76⁰

Question 2: In a △ABC, the internal bisectors of ∠B and ∠C meet at P and the external bisectors of ∠B and ∠C meet at Q. Prove that ∠BPC + ∠BQC = 180⁰.

Solution:

In triangle ABC,

BP and CP are internal bisector of ∠B and ∠C respectively

=> External ∠B = 180^o – ∠B

BQ and CQ are external bisector of ∠B and ∠C respectively.

=> External ∠C = 180 ^o – ∠C

In triangle BPC,

∠BPC + 1/2∠B + 1/2∠C = 180o

∠BPC = 180 ^o – (∠B + ∠C) …. (1)

In triangle BQC,

∠BQC + 1/2(180 ^o – ∠B) + 1/2(180 ^o – ∠C) = 180 ^o

∠BQC + 180 ^o – (∠B + ∠C) = 180 ^o

∠BPC + ∠BQC = 180 ^o [Using (1)]

Hence Proved.

Question 3: In figure, the sides BC, CA and AB of a △ABC have been produced to D, E and F respectively. If ∠ACD = 105⁰ and ∠EAF = 45⁰, find all the angles of the △ABC.

Solution:

∠BAC = ∠EAF = 45⁰ [Vertically opposite angles]

∠ACD = 180⁰ – 105⁰ = 75⁰ [Linear pair]

∠ABC = 105⁰ – 45⁰ = 60⁰ [Exterior angle property]

Question 4: Compute the value of x in each of the following figures:

(i)

Solution:

∠BAC = 180⁰ – 120⁰ = 60⁰ [Linear pair]

∠ACB = 180⁰ – 112⁰ = 68⁰ [Linear pair]

Sum of all angles of a triangle = 180⁰

x = 180⁰ − ∠BAC − ∠ACB

= 180⁰ − 60⁰ − 68⁰ = 52⁰

Answer: x = 52⁰

(ii)

Solution:

∠ABC = 180⁰ – 120⁰ = 60⁰ [Linear pair]

∠ACB = 180⁰ – 110⁰ = 70⁰ [Linear pair]

Sum of all angles of a triangle = 180⁰

x = ∠BAC = 180⁰ − ∠ABC − ∠ACB

= 1800 – 60⁰ – 70⁰ = 50⁰

Answer: x = 50⁰

(iii)

Solution:

∠BAE = ∠EDC = 52⁰ [Alternate angles]

Sum of all angles of a triangle = 180⁰

x = 180⁰ – 40⁰ – 52⁰ = 180⁰ − 92⁰ = 88⁰

Answer: x = 88⁰

(iv)

Solution:

CD is produced to meet AB at E.

∠BEC = 180⁰ – 45⁰ – 50⁰ = 85⁰ [Sum of all angles of a triangle = 180⁰]

∠AEC = 180⁰ – 85⁰ = 95⁰ [Linear Pair]

Now, x = 95⁰ + 35⁰ = 130⁰ [Exterior angle Property]

Answer: x = 130⁰

Question 5: In figure, AB divides ∠DAC in the ratio 1 : 3 and AB = DB. Determine the value of x.

Solution:

Let ∠BAD = y, ∠BAC = 3y

∠BDA = ∠BAD = y (As AB=DB)

Now,

∠BAD + ∠BAC + 108⁰ = 180⁰ [Linear Pair]

y + 3y + 108⁰ = 180⁰

4y = 72⁰

or y = 18⁰

Now, In ΔADC

∠ADC + ∠ACD = 108⁰ [Exterior Angle Property]

x + 18⁰ = 180⁰

x = 90⁰

Answer: x = 90⁰

Exercise VSAQs Page No: 9.21

Question 1: Define a triangle.

Solution: Triangle is a three-sided polygon that consists of three edges and three vertices. The most important property of a triangle is that the sum of the internal angles of a triangle is equal to 180 degrees.

Question 2: Write the sum of the angles of an obtuse triangle.

Solution: The sum of angles of obtuse triangle = 180°.

Question 3: In △ABC, if ∠B = 60⁰, ∠C = 80⁰ and the bisectors of angles ∠ABC and ∠ACB meet at point O, then find the measure of ∠BOC.

Solution:

∠B = 60⁰, ∠C = 80⁰ (given)

As per question:

∠OBC = 60⁰/2 = 30⁰ and

∠OCB = 80⁰/2 = 40⁰

In triangle BOC,

∠OBC + ∠OCB + ∠BOC = 180 ⁰

30⁰ + 40⁰ + ∠BOC = 180⁰

∠BOC = 110⁰

Question 4: If the angles of a triangle are in the ratio 2:1:3, then find the measure of smallest angle.

Solution:

Let angles of a triangles are 2x, x and 3x, where x is the smallest angle.

To find: measure of x.

As, Sum of angles of a triangle = 180⁰

2x + x + 3x = 180⁰

6x = 180⁰

x = 30⁰. Answer

Question 5: If the angles A, B and C of △ABC satisfy the relation B – A = C – B, then find the measure of ∠B.

Solution:

Sum of angles of a triangle = 180⁰

A + B + C = 180⁰ …(1)

B – A = C – B …(Given)

2B = C + A …(2)

(1) => 2B + B = 180⁰

3B =180⁰

Or B = 60⁰

RD Sharma Solutions for Chapter 9 Triangle and Its Angles

In the 9th chapter of Class 9 RD Sharma Solutions students will study important concepts listed below:

• Triangle Introduction
• Types of triangles
• Some important Theorems on triangles

Students can download all chapters solutions enlisted in RD Sharma class 9 textbook and start practising to score good marks.