Question

In the given figure if angle APB= 15 degree angle QAP = 60 degree, angle of depression of the object P from the point B is: *
15
30
45
60
​
Above given picture is asked in the question.

Answer 1

the b is 60 degree angel

Answer 2

Given : angle APB= 15 degree angle QAP = 60 degree,

To Find : angle of depression of the object P from the point B is:  

15°

30°

45°

60°

Solution:

in ΔAQP

∠QAP + ∠AQP + ∠QPA = 180°   ( Sum of angles of  a triangle )

=> 60° + 90° +  ∠QPA = 180°

=>  ∠QPA = 30°

∠QPB = ∠QPA + ∠APB

=> ∠QPB = 30° +  15°

=> ∠QPB =45°

Angle of depression from point B  = ∠QPB     alternate angles

=> Angle of depression from point B  to object P = 45°

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