Question

An observer, 1.5 m tall, is 28.5 m away from a 30 m high tower. Determine the angle of elevation of the top of the tower from the eye of the observer.

Answer 1

Answer:

The angle of elevation of the top of the tower from his eye is 45°.

Step-by-step explanation:

Given :  Height of the observer , DE = CD =  1.5 m.

Height of the tower, AD = 30m.

Distance between the observer and the tower , ED = BC = 28.5 m

Height of AC = AD - CD = 30 - 1.5 = 28.5 m  

In right angle ∆ ABC ,  

tan θ = P/B  

tan θ = AC/BC

tan θ = 28.5/28.5

tan θ = 1

θ = 45°  

Hence, the angle of elevation of the top of the tower from his eye is 45°.  

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Answer 2

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$\large\mathcal\red{solution}$

base=28.5 metres

tower height=30 metres

height of the man=1.5 metres

now...height of the tower with respect to the eye level of the man =(30-1.5)m=28.5 metres

now ...if theta be the angle of elevation.....

then ....

tan(theta)=height/base=(28.5/28.5)

tan(theta)=1

$theta = \tan {}^{ - 1} (1) = 45°$

so the angle of elevation is 45°

$\underline{\large\mathcal\red{hope\: this \: helps \:you......}}$