Question

A line segment AP stands at point P of a straight line BC such that ∠APB = 5x – 40° and ∠APC = .x+ 10°; find the value of x and angle APB.

Answer 1

AP stands on BC at P and

∠APB = 5x – 40°, ∠APC = x + 10°

(i) ∵ APE is a straight line

∠APB + ∠APC = 180°

⇒ 5x – 40° + x + 10° = 180°

⇒ 6x - 30°= 180°

⇒ 6x = 180° + 30° = 210°

x = `210^circ/6` = 35°

(ii) and ∠APB = 5x – 40° = 5 x 35° – 40°

= 175 ° – 40° = 135°