Question

plz solve 8, 9 and 10
and below answers is given..so plz..

Answer 1

(8)1) x+2x-15°+2x+35°+90° = 360°                   (Angles around a point)

5x+110° = 360°

5x = 360°-110°

5x = 250°

x = 250°/5 = 50°

2) 2y+2y+5y = 180°                 (Straight line)

9y = 180°

y = 20°

x = 5y                          (Vertically opposite angles)

x = 5*20°

x = 100°

(9)1) Since OD is the bisector of ∠BOC.

∠BOD = ∠COD = x

∠BOD + ∠COD + ∠AOC = 180°                   (Straight line)

x + x + 120°-x = 180°

x = 60°

2) ∠BOD = ∠COD = x                   (OD is bisector of ∠BOC)

x + x + 20° = 180°                           (Straight line)

2x = 160°

x = 80°

(10) Since ∠CPA:∠CPB = 7:5

∴∠CPA = 7x, ∠CPB = 5x

∠CPA + ∠CPB = 180°                               (Linear pair)

7x + 5x = 180°

12x = 180°

x = 15°

⇒∠CPA = 7x = 7*15 = 105°,∠CPB = 5x = 5*15 = 75°

Answer 2

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Answer 3

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