 # 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.

## 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 = 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

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 = 180o – ∠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

(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

(iii) Solution:

∠BAE = ∠EDC = 520 [Alternate angles]

Sum of all angles of a triangle = 1800

x = 1800 – 400 – 520 = 1800 − 920 = 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]

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

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

x + 180 = 1800

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

[Sum of angles of a triangle = 1800]

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

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.

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1. Ishitva Sinha

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