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.
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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 = 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
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
[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
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 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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