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# The angle of elevation of the top of the tower from a point on the ground is $${\sin ^{ - 1}}\left({\frac{3}{5}} \right).$$   If the point of observation is 20 meters away from the foot of the tower, what is the height of the tower?

A. 9 m

B. 18 m

C. 15 m

D. 12 m

Answer: Option C

### Solution(By Examveda Team)

Consider a right-angled triangle PQR as shown below.

Let QR = 3 and PR = 5 such that
$$\sin \theta = \frac{3}{5}\,\,\,\,\left[ {{\text{i}}{\text{.e}}{\text{.}},\theta = {{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right]$$
$$PQ = \sqrt {P{R^2} - Q{R^2}}$$     (∵ Pythagorean theorem)
\eqalign{ & = \sqrt {{5^2} - {3^2}} \cr & = 4 \cr}
$${\text{i}}{\text{.e}}{\text{.}},\,{\text{when}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right),$$     PQ : QR = 4 : 3 ......(eq : 1)

Now Let's solve the question. Let P be the point of observation and QR be the tower as shown in the below diagram.

$${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$     and PQ = 20 m
We know that PQ : QR = 4 : 3 (from eq : 1)
\eqalign{ & {\text{i}}{\text{.e}}{\text{.}},\,20:QR = 4:3 \cr & \Rightarrow 20 \times 3 = QR \times 4 \cr & \Rightarrow QR = 15\,{\text{m}} \cr}
i.e. height of the tower = 15 m

Solution 2

Let P be the point of observation and QR be the tower.
$${\text{Given}}\,{\text{that}}\,\theta = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right)$$     and PQ = 20 m
Let the height of the tower, QR = h and PR = x
\eqalign{ & {\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta PQR, \cr & \sin \theta = \frac{{QR}}{{PR}} \cr & \Rightarrow \sin \left[ {{{\sin }^{ - 1}}\left( {\frac{3}{5}} \right)} \right] = \frac{h}{x} \cr & \Rightarrow \frac{3}{5} = \frac{h}{x} \cr & \Rightarrow x = \frac{{5h}}{3}\,.....\left( {eq:1} \right) \cr}
From Pythagorean theorem, we have
\eqalign{ & P{Q^2} + Q{R^2} = P{R^2} \cr & {20^2} + {h^2} = {x^2} \cr}
$${20^2} + {h^2} = {\left( {\frac{{5h}}{3}} \right)^2}$$   (∵ Substituted the value of x from eq:1)
\eqalign{ & {20^2} + {h^2} = \frac{{25{h^2}}}{9} \cr & \frac{{16{h^2}}}{9} = {20^2} \cr & \frac{{4h}}{3} = 20 \cr & h = \frac{{3 \times 20}}{4} = 15{\text{m}} \cr}
i.e. height of the tower = 15 m

This Question Belongs to Arithmetic Ability >> Height And Distance

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