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RS Aggarwal Solutions Class 9 Maths Chapter 7 Lines and Angles
1: Define the following terms:
(i) Angle
When two rays originated from same point, then an angle is formed.
(ii) Interior of an angle
The area between two rays which is originated from same point, is called Interior of an angle.
(iii) Obtuse angle
An angle which is measured more than 900 but less than 1800 is called Obtuse angle.
(iv) Reflex angle
An angle which is measured more than 1800 but less than 3600 is called Reflex angle.
(v) Complementary angle
Two angle are said to be complementary if, the sum of both angles are 900 .
(vi) Supplementary angle
Two angle are said to be supplementary, if the sum of both angles are 1800.
2: Find the complement of each of the following angles:
(i) 550 (ii) 160 (iii) 900 (iv) 2/3 of a right angle
Complement of 550 = 900 – 550 = 350
Complement of 160 = 900 – 160= 740
Complement of 900 = 900 – 900 = 00
Complement of 2/3 of a right angle
= 900 – ( 900 × 2/3) = 900 – 600 =300
3: Find the supplementary of each of the following angles:
(i) 420 (ii) 900 (iii) 1240 (iv) 3/5 of a right angle
Supplementary of 420 = 1800 – 420 =1380
Supplementary of 900 = 1800 – 900 = 900
Supplementary of 1240 = 1800 – 1240 = 560
Supplementary of 3/5 of a right angle
= 1800 – ( 900 × 3/5) = 1800 – 540= 1260
4: Find the measure of an angle which is
(i) equal to its complement,
Let the angle is x0 , then
⇒ x0 = 900 – x0
⇒ 2x0 = 900
⇒ x0 = 450
Hence the angle is 450 which is equal to its complement.
(ii) equal to its supplement
Let the angle is x0 then
⇒ x0 = 1800 – x0
⇒ 2x0 = 1800
⇒ x0 = 900
Hence the angle is 900 which is equal to its supplement.
5: Find the measure of an angle which is 360 more than its complement.
Let the angle is x0 , then its complement angle will be 900 – x0
According to question,
⇒ x0 – 360 = 900 – x0
⇒ 2x0 = 900 + 360 = 1260
⇒ x0 = 630
Hence the angle is 630 .
6: Find the measure of an angle which is 300 less than its supplement.
Let the angle is x0, then its supplement angle will be 1800 – x0
According to question,
⇒ x0 +300 = 1800 – x0
⇒ 2x0 = 1800 – 300 = 1500
⇒ x0 = 750
Hence the angle is 750.
7: Find the angle which is four times its complement.
Let the angle is x0, then its complement angle will be 900 – x0
According to question,
⇒ x0 = 4(900 – x0 )
⇒ x0 = 3600 – 4x0
⇒ 5x0 = 3600
⇒ x0= 720
Hence the angle is 720
8: Find the angle which is five times its supplement.
Let the angle is x0, then its supplementary angle will be 1800 – x0
According to question,
⇒ x0 = 5(1800 – x0 )
⇒ x0 = 9000 – 5x0
⇒ 6x0 = 9000
⇒ x0 = 1500
Hence the angle is 1500 .
9: Find the angle whose supplement is four times its complement.
Let the angle is x0, then its supplement angle will be 1800 – x0
and complement will be 900 – x0
According to question,
⇒ (1800 – x0 ) = 4(900 – x0 )
⇒ 1800 – x0 = 3600 – 4x0
⇒ 4x0 – x0 = 3600 – 1800 = 1800
⇒ 3x0=1800
⇒ x0 = 600
10: Find the angle whose complement is one third of its supplement.
Let the angle is x0 , then its supplement angle will be 1800 – x0
and complement will be 900 – x0
According to question,
⇒ (900 – x0) = 1/3 (1800 – x0)
⇒2700 – 3x0= 1800 – x0
⇒ – 3x0 + x0 = 1800 – 2700 = –900
⇒ – 2x0 = – 900
⇒ x0 = 450
Hence the angle is 450
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