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: c
Explanation: Consider a right-angled triangle PQR
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)

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
We know that PQ : QR = 4 : 3 (from eq : 1)
$$\eqalign{ & {\text{i}}{\text{.e}}{\text{.}},\,20:QR = 4:3 \cr & \Rightarrow 20 \times 3 = QR \times 4 \cr & \Rightarrow QR = 15\,{\text{m}} \cr} $$
Height of the tower = 15 m