# A tower is $5\sqrt 3 $ meter high. Find the angle of elevation of its top from a point 5 meter away from its foot.

Last updated date: 24th Mar 2023

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Answer

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Hint- Here, we will be making diagram according to the problem statement and then we will use the formula for tangent trigonometric function i.e, $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$ in order to evaluate the value for the angle of elevation (i.e., $\theta $).

Complete step-by-step answer:

Given, height of the tower AB = $5\sqrt 3 $ meter

Let point C be a point which is 5 meter away from the foot of the tower AB (i.e., point B).

Let us suppose that the angle of elevation of the top of the tower (i.e., point A) from point C is $\theta $.

As we know that in any right angled triangle, $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}{\text{ }} \to {\text{(1)}}$

In the right angled triangle ABC (right angled at vertex B), side AB is the perpendicular, side BC is the base and side AC is the hypotenuse.

Using formula given by equation (1) for triangle ABC, we get

$

\tan \theta = \dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \dfrac{{5\sqrt 3 }}{5} \\

\Rightarrow \tan \theta = \sqrt 3 {\text{ }} \to {\text{(2)}} \\

$

Also we know that $\tan {60^0} = \sqrt 3 {\text{ }} \to {\text{(3)}}$

Clearly, the RHS of both the equations (2) and (3) is the same so the LHS of both these equations will also be equal.

i.e., $

\tan \theta = \tan {60^0} \\

\Rightarrow \theta = {60^0} \\

$

Therefore, the required angle of elevation of the top of the tower from a point 5 meter away from the foot of the tower is ${60^0}$.

Note- In any right angled triangle, the hypotenuse is the side opposite to ${90^0}$ (in this case the right angle is at B and the side opposite to vertex B is AC), the perpendicular is the side opposite to the considered angle $\theta $ (in this case the perpendicular is AB) and the base is the remaining side (in this case base is BC).

Complete step-by-step answer:

Given, height of the tower AB = $5\sqrt 3 $ meter

Let point C be a point which is 5 meter away from the foot of the tower AB (i.e., point B).

Let us suppose that the angle of elevation of the top of the tower (i.e., point A) from point C is $\theta $.

As we know that in any right angled triangle, $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}{\text{ }} \to {\text{(1)}}$

In the right angled triangle ABC (right angled at vertex B), side AB is the perpendicular, side BC is the base and side AC is the hypotenuse.

Using formula given by equation (1) for triangle ABC, we get

$

\tan \theta = \dfrac{{{\text{AB}}}}{{{\text{BC}}}} = \dfrac{{5\sqrt 3 }}{5} \\

\Rightarrow \tan \theta = \sqrt 3 {\text{ }} \to {\text{(2)}} \\

$

Also we know that $\tan {60^0} = \sqrt 3 {\text{ }} \to {\text{(3)}}$

Clearly, the RHS of both the equations (2) and (3) is the same so the LHS of both these equations will also be equal.

i.e., $

\tan \theta = \tan {60^0} \\

\Rightarrow \theta = {60^0} \\

$

Therefore, the required angle of elevation of the top of the tower from a point 5 meter away from the foot of the tower is ${60^0}$.

Note- In any right angled triangle, the hypotenuse is the side opposite to ${90^0}$ (in this case the right angle is at B and the side opposite to vertex B is AC), the perpendicular is the side opposite to the considered angle $\theta $ (in this case the perpendicular is AB) and the base is the remaining side (in this case base is BC).

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