RD Sharma Solutions for Class 9 Maths are given here for Chapter 8 – Lines and Angles. In order to make concepts easier for students, we have provided RD Sharma Class 9 Maths Chapter 8 solutions to help them go through several solved exercises and, at the same time, practise the questions. RD Sharma Solutions for Class 9 can help students discover new ways to solve difficult problems. They can download RD Sharma Class 9 solutions and practise offline.
RD Sharma Solutions for Class 9 Maths Chapter 8 Lines and Angles Exercise 8.1
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 8 Lines and Angles Exercise 8.1 Page Number 8.7
Question 1: Write the complement of each of the following angles:
(i)200
(ii)350
(iii)900
(iv) 770
(v)300
Solution:
(i) The sum of an angle and its complement = 900
Therefore, the complement of 200 = 900 – 200 = 700
(ii) The sum of an angle and its complement = 900
Therefore, the complement of 35° = 90° – 35° = 55
(iii) The sum of an angle and its complement = 900
Therefore, the complement of 900 = 900 – 900 = 00
(iv) The sum of an angle and its complement = 900
Therefore, the complement of 770 = 90° – 770 = 130
(v) The sum of an angle and its complement = 900
Therefore, the complement of 300 = 900 – 300 = 600
Question 2 : Write the supplement of each of the following angles:
(i) 540
(ii) 1320
(iii) 1380
Solution:
(i) The sum of an angle and its supplement = 1800.
Therefore supplement of angle 540 = 1800 – 540 = 1260
(ii) The sum of an angle and its supplement = 1800.
Therefore supplement of angle 1320 = 1800 – 1320 = 480
(iii) The sum of an angle and its supplement = 1800.
Therefore supplement of angle 1380 = 1800 – 1380 = 420
Question 3: If an angle is 280 less than its complement, find its measure?
Solution:
Let the measure of any angle is ‘ a ‘ degrees
Thus, its complement will be (90 – a) 0
So, the required angle = Complement of a – 28
a = ( 90 – a ) – 28
2a = 62
a = 31
Hence, the angle measured is 310.
Question 4 : If an angle is 30° more than one half of its complement, find the measure of the angle?
Solution:
Let an angle measured by ‘ a ‘ in degrees
Thus, its complement will be (90 – a) 0
Required Angle = 300 + complement/2
a = 300 + ( 90 – a ) 0 / 2
a + a/2 = 300 + 450
3a/2 = 750
a = 500
Therefore, the measure of required angle is 500.
Question 5 : Two supplementary angles are in the ratio 4:5. Find the angles?
Solution:
Two supplementary angles are in the ratio 4:5.
Let us say, the angles are 4a and 5a (in degrees)
Since angle are supplementary angles;
Which implies, 4a + 5a = 1800
9a = 1800
a = 200
Therefore, 4a = 4 (20) = 800 and
5(a) = 5 (20) = 1000
Hence, required angles are 80° and 1000.
Question 6 : Two supplementary angles differ by 480. Find the angles?
Solution: Given: Two supplementary angles differ by 480.
Consider a0 be one angle then its supplementary angle will be equal to (180 – a) 0
According to the question;
(180 – a ) – x = 48
(180 – 48 ) = 2a
132 = 2a
132/2 = a
Or a = 660
Therefore, 180 – a = 1140
Hence, the two angles are 660 and 1140.
Question 7: An angle is equal to 8 times its complement. Determine its measure?
Solution: Given: Required angle = 8 times of its complement
Consider a0 be one angle then its complementary angle will be equal to (90 – a) 0
According to the question;
a = 8 times of its complement
a = 8 ( 90 – a )
a = 720 – 8a
a + 8a = 720
9a = 720
a = 80
Therefore, the required angle is 800.
RD Sharma Solutions for Class 9 Maths Chapter 8 Lines and Angles Exercise 8.1
Class 9 Maths Chapter 8 Lines and Angles Exercise 8.1 is based on the following topics and subtopics:
- Angles
- Interior of an angle
- Exterior of an angle
- Congruent angles
- Measure of an angle
- Axioms and theorems
- Types of angles
- Right Angle
- Acute Angle
- Obtuse Angle
- Straight Angle
- Reflex Angle
- Complementary Angles
- Supplementary Angles
- Angle bisector
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