RD SHARMA Solutions for Class 9 Maths Chapter 10 - Lines and Angles
Chapter 10 - Lines and Angles Exercise Ex. 10.1
(i) 54°
Since, the sum of an angle and its supplement is 180°
∴Its supplement will be 180° - 54° = 126°.
(ii) 132°
Since, the sum of an angle and its supplement is 180°
∴Its supplement will be 180° - 132° = 48°.
(iii) 138°
Since, the sum of an angle and its supplement is 180°
∴Its supplement will be 180° - 138° = 42°.
Let the measure of the angle be xo.
Its complement will be (90o - xo) and its supplement will be (180o - xo).
Supplement of thrice of the angle = (180o - 3xo)
According to the given information:
(90o - xo) = (180o - 3xo)
3x - x = 180 - 90
2x = 90
x = 45
Thus, the measure of the angle is 45o.
The measure of the angle is 45o
Chapter 10 - Lines and Angles Exercise Ex. 10.2

Chapter 10 - Lines and Angles Exercise Ex. 10.3
(i) True
(ii) False
(iii) False
(iv) True
(i) obtuse.
(ii) 180o
(iii) uncommon
Chapter 10 - Lines and Angles Exercise Ex. 10.4
Consider the angles AOB and ACB.
Chapter 10 - Lines and Angles Exercise 10.51
Chapter 10 - Lines and Angles Exercise 10.52
Chapter 10 - Lines and Angles Exercise 10.53
Chapter 10 - Lines and Angles Exercise 10.54
Chapter 10 - Lines and Angles Exercise 10.55
Chapter 10 - Lines and Angles Exercise 10.56
Correct option (c)
Let one angle be θ
Then, its complementary = 90 - θ
According to question,
2θ = 3(90 - θ)
5θ = 270
θ = 54°
Then, 90 - θ° = 36°
Hence, the smaller angle is 36°.
Hence, correct option is (c).
Chapter 10 - Lines and Angles Exercise 10.57
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