RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1 (Updated for 2024)

RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1: Class 9 Maths is not actually difficult if you are studying it from the RD Sharma Solutions Class 9 Maths. You can easily clear your doubts and learn the concepts in the RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1. You can download the PDF from the link given in this blog. To know more, read the whole blog. 

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RD Sharma Class 9 Solutions Chapter 10 Exercise 10.1

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Question 1.
Write the complement of each of the following angles:
(i) 20°
(ii) 35°
(iii) 90°
(iv) 77°
(v) 30°
Solution:
We know that two angles complement each other if their sum is 90°. Therefore,
(i) Complement of 20° is (90° – 20°) = 70°
(ii) Complement of 35° is (90° – 35°) = 55°
(iii) Complement of 90° is (90° – 90°) = 0°
(iv) Complement of 77° is (90° – 77°) = 13°
(v) Complement of 30° is (90° – 30°) = 60°

Question 2.
Write the supplement of each of the following angles:
(i) 54°
(ii) 132°
(iii) 138°
Solution:
We know that two angles are supplementary to each other if their sum is 180°. Therefore,
(i) Supplement of 54° is (180° – 54°) = 126°
(ii) Supplement of 132° is (180° – 132°) = 48°
(iii) Supplement of 138° is (180° – 138°) = 42°

Question 3.
If an angle is 28° less than its complement, find its measure.
Solution:
Let the required angle = x, then
Its complement = x + 28°
∴  x + x + 28° = 90° ⇒  2x = 90° – 28° = 62°
∴ x = 62∘2  = 31°
∴ Required angle = 31°

Question 4.
If an angle is 30° more than one-half of its complement, find the measure of the angle.
Solution:
Let the measure of the required angle = x
∴  Its complement =  90° – x
∴  x = 12 (90° – x) + 30°
2x = 90° – x + 60°
⇒ 2x + x = 90° + 60°
⇒  3x = 150°
⇒ x =  150∘3  = 50°
∴ Required angle = 50°

Question 5.
Two supplementary angles are in the ratio 4: 5. Find the angles.
Solution:
Ratio in two supplementary angles = 4 : 5
Let first angle = 4x
The second angle = 5x
∴  4x + 5x = 180
⇒  9x = 180°
∴ x  = 180∘9 = 20°
∴  First angle = 4x = 4 x 20° = 80°
and second angle = 5x
= 5 x 20° = 100°

Question 6.
Two supplementary angles differ by 48°. Find the angles.
Solution:
Let first angle = x                        ”
The second angle = x + 48°
∴  x + x + 48° = 180°⇒  2x + 48° = 180°
⇒  2x = 180° – 48° = 132°
x= 132∘2 =66°
∴  First angle = 66°
and second angle = x + 48° = 66° + 48° = 114°
∴ Angles are 66°, 114°

Question 7.
An angle is equal to 8 times its complement. Determine its measure.
Solution:
Let the required angle = x
Then its complement angle = 90° – x
∴ x = 8(90° – x)
⇒ x = 720° – 8x ⇒  x + 8x = 720°
⇒ 9x = 720° ⇒ x =  720∘9 = 80°
∴  Required angle = 80°

Question 8.
If the angles (2x – 10)° and (x – 5)° are complementary angles, find x.
Solution:
First complementary angle = (2x – 10°) and second = (x – 5)°
∴ 2x – 10° + x – 5° = 90°
⇒ 3x – 15° = 90° ⇒  3x = 90° + 15° = 105°
∴ x = 105∘3 = 35°
∴  First angle = 2x – 10° = 2 x 35° – 10°
= 70° – 10° = 60°
and second angle = x – 5 = 35° – 5 = 30°

Question 9.
If an angle differs from its complement by 10°, find the angle.
Solution:
Let required angle = x°
Then its complement angle = 90° – x°
∴ x – (90° – x) = 10
⇒  x – 90° + x = 10°⇒  2x = 10° + 90° = 100° 100°
⇒ x =  100∘2 = 50°
∴ Required angle = 50°

Question 10.
If the supplement of an angle is two-thirds of itself Determine the angle and its supplement.
Solution:
Let required angle = x
Then its supplement angle = 180° – x
∴  (180°-x)= 23x
540° – 3x = 2x ⇒ 2x + 3x = 540°
⇒ 5x = 540°⇒  x = 540∘5 = 108°
-. Supplement angle = 180° – 108° = 72°

Question 11.
An angle is 14° more than its complementary angle. What is its measure?
Solution:
Let required angle = x
Then its complementary angle = 90° – x
∴  x + 14° = 90° – x
x + x = 90° – 14° ⇒  2x = 76°
⇒ x =  76∘2 = 38°
∴  Required angle = 38°

Question 12.
The measure of an angle is twice the measure of its supplementary angle. Find its measure.
Solution:
Let the required angle = x
∴  Its supplementary angle = 180° – x
∴  x = 2(180°-x) = 360°-2x
⇒  x + 2x = 360°
⇒ 3x = 360°
⇒  x = 360∘3 = 120°
∴  Required angle = 120°

Question 13.
If the complement of an angle is equal to the supplement of the thrice of it. Find the measure of the angle.
Solution:
Let required angle = x
Then its complement angle = 90° – x
and supplement angle = 180° – x
∴  3(90° – x) = 180° – x
⇒ 270° – 3x = 180° – x
⇒270° – 180° = -x + 3x => 2x = 90°
⇒ x = 45°
∴  Required angle = 45°

Question 14.
If the supplement of an angle is three times its complement, find the angle.
Solution:
Let required angle = x
Then its complement = 90°-  x
and supplement = 180° – x
∴  180°-x = 3(90°-x)
⇒  180° – x = 270° – 3x
⇒  -x + 3x = 270° – 180°
⇒ 2x = 90° ⇒ x = 90∘2 =45°
∴ Required angle = 45°

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