RS Aggarwal And Veena Aggarwal Class 9 Math Seventh Chapter Lines and Angles Exercise 7A Solution

RS Aggarwal And Veena Aggarwal Class 9 Math Seventh Chapter Lines and Angles Exercise 7A Solution

EXERCISE 7A

(1) (i) Angle – When two rays originate from the same end point, then an angle is formed.

(ii) Interior of an angle – The interior of ∠BAC is the set of all points in its plane which lie on the same side of AB as C and also on the same side of AC as B.

(iii) Obtuse angle – An angle whose measure is more than 90o but less than 180o is called an obtuse angle.

(iv) Reflex angle – An angle whose measure is more than 180o but less than 360o is called a reflex angle.

(v) Complementary angles – Two angles are said to be complementary, if the sum of their measures is 90o.

(vi) Supplementary angles – Two angles are said to be supplementary if the sum of their measures is 180o.

(2) (i) Let the measure of the required angle be xo.

As sum of two complementary angles is = 90o.

∴ x + 55 = 90

⇒ x = 90 – 55

⇒ x = 35

(ii) Let the measure of the required angle be xo.

As sum of two complementary angles is = 90o.

∴ x + 16 = 90

⇒ x = 90 – 16

⇒ x = 74

(iii) Let the measure of the required angle be xo.

As sum of two complementary angles is = 90o.

∴ x + 90 = 90

⇒ x = 90 – 90

⇒ x = 0

(iv) Let the measure of the required angle be xo.

As sum of two complementary angles is = 90o.

(3) (i) Let the measure of the required angle be xo.

As sum of two complementary angles is = 180o.

∴ x + 42 = 180

⇒ x = 180 – 42

⇒ x = 138

(ii) Let the measure of the required angle be xo.

As sum of two complementary angles is = 180o.

∴ x + 90 = 180

⇒ x = 180 – 90

⇒ x = 90

(iii) Let the measure of the required angle be xo.

As sum of two complementary angles is = 180o.

∴ x + 124 = 180

⇒ x = 180 – 124

⇒ x = 56

(iv) Let the measure of the required angle be xo.

As sum of two complementary angles is = 180o.

(4) (i) Let the measure of the required angle be xo.

∴ x + x = 90

⇒ 2x = 90

⇒ x = 45

(ii) Let the measure of the required angle be xo.

∴ x+ x = 180

⇒ 2x = 180

⇒ x = 90

(5) Let the measure of the required angle be xo.

∴ x – (90 – x) = 36

⇒ x – 90 + x = 36

⇒ 2x = 36 + 90 = 126

⇒ x = 63

(6) Let the measure of the required angle be xo.

∴ (180 – x) – x = 30

⇒ 180 – 2x = 30

⇒ – 2x = 30 – 180

⇒ – 2x = – 150

⇒ x = 75

(7) Let the measure of the required angle be xo.

∴ x = 4(90 – x)

⇒ x = 360 – 4x

⇒ x + 4x = 360

⇒ 5x = 360

⇒ x = 72

(8) Let the measure of the required angle be xo.

∴ x = 5(180 – x)

⇒ x = 900 – 5x

⇒ x + 5x = 900

⇒ 6x = 900

⇒ x = 150

(9) Let the measure of the required angle be xo.

∴ (180 – x) = 4(90 – x)

⇒ 180 – x = 360 – 4x

⇒ – x + 4x = 360 – 180

⇒ 3x = 180

⇒ x = 60

(10) Let the measure of the required angle be xo.

∴ (90 – x) = (180 – x)

⇒ 270 – 3x = 180 – x

⇒ – x + 3x = 270 – 180

⇒ 2x = 90

⇒ x = 45

(11) Let the measure of two complementary angles be 4xo and 5xo.

∴ 4x + 5x = 90

⇒ 9x = 90

⇒ x = 10

Therefore, the two complementary angles are (4 × 10) = 40o and (5 × 10) = 50o.

(12) The complementary angles are

∴ (2x – 5) + (x – 10) = 90

⇒ 2x – 5 + x – 10 = 90

⇒ 3x – 15 = 90

⇒ 3x = 90 + 15 = 105

⇒ x = 35

 

For more exercise solution, Click below –

Leave a Reply

Your email address will not be published. Required fields are marked *