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

RD Sharma Solutions Class 9 Maths Chapter 9 – Free PDF Download

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. The solutions are so apt that students who practice on a regular basis undoubtedly secure high marks in their academics. 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 a plane figure formed by three non-parallel lines in a plane, called a triangle. A triangle is a 2D geometrical figure consisting of three edges and three vertices. 

The three angles are 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: Acute Triangle, Right Angle and Obtuse Triangle. For a better understanding of concepts, students can download the solutions in PDF whenever required. RD Sharma Solutions are the best material for students who aspire to become proficient in Mathematics. The solutions are designed based on the latest CBSE syllabus as per the student’s intelligence quotient.

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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 = 180^o

∠BPC = 180 ^o – 1/2(∠B + ∠C) …. (1)

In triangle BQC,

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

∠BQC + 180 ^o – 1/2(∠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:

$\angle DAC=180°-108°=72°$

∠BAC/∠DAB​=1/3

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 Class 9 Maths 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.