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The measure of angle of elevation of top of tower 75{3} m high

Question

The measure of angle of elevation of top of tower $753$ m high from a point at a distance of 75 m from foot of tower in a horizontal plane is :

3 0^{\circ}

6 0^{\circ}

9 0^{\circ}

4 5^{\circ}

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The measure of angle of elevation of top of tower

753

m high from a point at a distance of 75 m from foot of tower in a horizontal plane is :

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Text solutionVerified

Let the angle of deviation be

θ .

△ A CB,

\Rightarrow tan θ = \frac{A B}{BC} = \frac{75 3}{75}

\Rightarrow tan θ = 3

\Rightarrow θ = 6 0^{\circ}

Hence, the answer is

6 0^{\circ} .

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Question Text	The measure of angle of elevation of top of tower $753$ m high from a point at a distance of 75 m from foot of tower in a horizontal plane is :
Answer Type	Text solution:1
Upvotes	154