The angle of elevation of the top of the tower from a point on the ground is ^ - 1 (3/5 ). If the point of observation is 20 meters away from the foot of the tower, what is the height of the tower?

Answer: Option C

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)
$$\begin{array}{rl}& =\sqrt{5^2-3^2}\\ & =4\end{array}$$
$$\text{i}\text{.e}\text{.},\text{when}\theta ={\sin }^{-1}\left(\frac{3}{5}\right),$$     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)
$$\begin{array}{rl}& \text{i}\text{.e}\text{.},20:QR=4:3\\ & \Rightarrow 20\times 3=QR\times 4\\ & \Rightarrow QR=15\text{m}\end{array}$$
i.e. height of the tower = 15 m

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