Question

what is the measure of the angle of elevation of the top of a tower 57√3m high from a point 57 m from the foot of the tower?
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Answer 1

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Answer 2

Given : 
Height of the tower                                            = 57√3m 
Distance from the point to the foot of the tower = 57 

To Find : 
The angle of elevation 

Solution : 

Let us take the AB = height of tower = 57√3 
                        BC = distance between the point and foot of the tower = 57 

Tan theta = Opposite side / Adjacent side 
⇒ tan theta = AB/BC 
⇒ tan theta = 57√3  / 57 
( 57 cancels 57) 
⇒ tan theta = √3 
(∵ tan 60° = √3) 
∴ theta = 60° 

∴ the angle of elevation is 60° of the top of a tower 57√3  high from a point 57m from the foot of the tower