A vertical tower is $20$ m high. A man at some distance from the tower knows that the cosine of the angle of the elevation of the top of the tower is $0.5$ . He is standing from the foot of the tower at a distance of:

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Solution

Let height be $^{'} h^{'} = 20 m$
And the distance of the man from tower be $^{'} d^{'}$

According to question angle of elevation is
$c o s θ = 0.5$
=> $θ = 60 °$

From figure ,
$t a n θ = \frac{20}{d}$
=> $\sqrt 3 = \frac{20}{d}$

=> $d = \frac{20}{\sqrt 3} m$

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A vertical tower is 20 m high. A man at some distance from the tower knows that the cosine of the angle of the elevation of the top of the tower is 0.5. He is standing from the foot of the tower at a distance of:

Let height be ′h′=20mAnd the distance of the man from tower be ′d′According to question angle of elevation iscosθ=0.5=> θ=60°From figure ,tanθ=20d=>√3=20d=> d=20√3m

A vertical tower is $20$ m high. A man at some distance from the tower knows that the cosine of the angle of the elevation of the top of the tower is $0.5$ . He is standing from the foot of the tower at a distance of:

Let height be $^{'} h^{'} = 20 m$
And the distance of the man from tower be $^{'} d^{'}$

According to question angle of elevation is
$c o s θ = 0.5$
=> $θ = 60 °$

From figure ,
$t a n θ = \frac{20}{d}$
=> $\sqrt 3 = \frac{20}{d}$

=> $d = \frac{20}{\sqrt 3} m$