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in Mathematics by (61.9k points)

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.

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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 = (210/6)° 

= 35°

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

= 175 ° –  40° 

= 135°

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