RD Sharma Class 9 Solutions Chapter 9 is provided here for students to clear all their doubts regarding triangles as well as problems based on this topic. RD Sharma Solutions 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.

## Download PDF of RD Sharma Solutions for Class 9 Chapter 9 Triangle and its Angles

### Access Answers to Maths RD Sharma Chapter 9 Triangle and its Angles

Exercise 9.1 Page No: 9.9

**Question 1: In a ΔABC, if ∠A = 55 ^{0}, ∠B = 40^{0}, find ∠C.**

**Solution:**

Given: ∠A = 55^{0}, ∠B = 40^{0}

We know, sum of all angles of a triangle is 180^{0}

∠A + ∠B + ∠C = 180^{0}

55^{0 }+ 40^{0 }+ ∠C=180^{0}

95^{0 }+ ∠C = 180^{0}

∠C = 180^{0} − 95^{0}

∠C = 85^{0}

**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 = 180^{0}

x + 2x + 3x = 180^{0}

6x = 180^{0}

x = 180^{0}/6

x = 30^{0}

Answer:

x = 30^{0}

2x = 2(30)^{0} = 60^{0}

3x = 3(30)^{ 0} = 90^{0}

**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 = 180^{0}

(x − 40)^{0} + (x − 20)^{ 0 }+ (1/2 x − 10)^{ 0} = 180^{0}

5/2 x – 70^{0} = 1800

5/2 x = 180^{0} + 70^{0}

5x = 2(250)^{ 0}

x = 500^{0}/5

x = 100^{0}

**Question 4: The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10 ^{0}, find the three angles.**

**Solution:**

The difference between two consecutive angles is 10^{0} (given)

Let x, x + 10^{0}, x + 20^{0} be the consecutive angles

x + x + 10^{0} + x + 20^{0} = 180^{0}

3x + 30^{0} = 180^{0}

3x = 180^{0}– 30^{0}

3x = 150^{0}

or x = 50^{0}

Again,

x + 10^{0} = 50^{0} + 10^{0} = 60^{0}

x+20^{0} = 50^{0} + 20^{0} = 70^{0}

Answer: Three angles are 50^{0},60^{0} and 70^{0}.

**Question 5: Two angles of a triangle are equal and the third angle is greater than each of those angles by 30 ^{0}. 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 30^{0}. (Given)

Let x, x, x + 30^{0} be the angles of a triangle.

Sum of all angles in a triangle = 180^{0}

x + x + x + 30^{0} = 180^{0}

3x + 30^{0} = 180^{0}

3x = 150^{0}

or x = 50^{0}

And x + 30^{0} = 50^{0} + 30^{0} = 80^{0}

Answer: Three angles are 50^{0}, 50^{0} and 80^{0}.

**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 90^{0}

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

z = x + y …(1)

Sum of all angles of a triangle = 180^{0}

x + y + z = 180^{0}

z + z = 180^{0} (Using equation (1))

2z = 180^{0}

z = 90^{0} (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 104 ^{0} and 136^{0}. Find all the angles of the triangle.**

**Solution:**

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

Find ∠ABC:

∠ABC + ∠ABE = 180^{0 } [Linear pair]

∠ABC + 136^{0 }= 180^{0}

∠ABC = 44^{0}

Find ∠ACB:

∠ACB + ∠ACD = 180^{0} [Linear pair]

∠ACB + 104^{0} = 180^{0}

∠ACB = 76^{0}

Now,

Sum of all angles of a triangle = 180^{0}

∠A + 44^{0} + 76^{0} = 180^{0}

∠A = 180^{0 }− 44^{0 }−76^{0}

∠ A = 60^{0}

Answer: Angles of a triangle are ∠ A = 60^{0}, ∠B = 44^{0} and ∠C = 76^{0}

**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 = 180 ^{0}.**

**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 = 105 ^{0} and ∠EAF = 45^{0}, find all the angles of the △ABC.**

**Solution:**

∠BAC = ∠EAF = 45^{0} [Vertically opposite angles]

∠ACD = 180^{0} – 105^{0} = 75^{0} [Linear pair]

∠ABC = 105^{0} – 45^{0} = 60^{0} [Exterior angle property]

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

**(i)**

**Solution:**

∠BAC = 180^{0} – 120^{0} = 60^{0} [Linear pair]

∠ACB = 180^{0} – 112^{0} = 68^{0} [Linear pair]

Sum of all angles of a triangle = 180^{0}

x = 180^{0 }− ∠BAC − ∠ACB

= 180^{0 }− 60^{0 }− 68^{0 }= 52^{0}

Answer: x = 52^{0}

**(ii)**

**Solution:**

∠ABC = 180^{0} – 120^{0} = 60^{0} [Linear pair]

∠ACB = 180^{0} – 110^{0} = 70^{0} [Linear pair]

Sum of all angles of a triangle = 180^{0}

x = ∠BAC = 180^{0} − ∠ABC − ∠ACB

= 1800 – 60^{0} – 70^{0} = 50^{0}

Answer: x = 50^{0}

**(iii)**

**Solution:**

∠BAE = ∠EDC = 52^{0} [Alternate angles]

Sum of all angles of a triangle = 180^{0}

x = 180^{0} – 40^{0} – 52^{0} = 180^{0 }− 92^{0} = 88^{0}

Answer: x = 88^{0}

**(iv) **

**Solution:**

CD is produced to meet AB at E.

∠BEC = 180^{0} – 45^{0} – 50^{0} = 85^{0} [Sum of all angles of a triangle = 180^{0}]

∠AEC = 180^{0} – 85^{0} = 95^{0} [Linear Pair]

Now, x = 95^{0} + 35^{0} = 130^{0} [Exterior angle Property]

Answer: x = 130^{0}

**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 + 108^{0} = 180^{0} [Linear Pair]

y + 3y + 108^{0} = 180^{0}

4y = 72^{0}

or y = 18^{0}

Now, In ΔADC

∠ADC + ∠ACD = 108^{0} [Exterior Angle Property]

x + 18^{0 }= 180^{0}

x = 90^{0}

Answer: x = 90^{0}

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 = 60**^{0}**, ∠C = 80**^{0}** and the bisectors of angles ∠ABC and ∠ACB meet at point O, then find the measure of ∠BOC.**

**Solution**:

∠B = 60^{0}, ∠C = 80^{0} (given)

As per question:

∠OBC = 60^{0}/2 = 30^{0} and

∠OCB = 80^{0}/2 = 40^{0}

In triangle BOC,

∠OBC + ∠OCB + ∠BOC = 180 ^{0}

^{0}]

30^{0} + 40^{0} + ∠BOC = 180^{0}

∠BOC = 110^{0}

**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 = 180^{0}

2x + x + 3x = 180^{0}

6x = 180^{0}

x = 30^{0}. 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 = 180^{0}

A + B + C = 180^{0} …(1)

B – A = C – B …(Given)

2B = C + A …(2)

(1) => 2B + B = 180^{0}

3B =180^{0}

Or B = 60^{0}

## RD Sharma Solutions For Class 9 Maths Chapter 9 Exercises:

Get detailed solutions for all the questions listed under below exercises:

## RD Sharma Solutions for Chapter 9 Triangle and its Angles

In this chapter 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.

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