[15 m] The angle of elevation of the top of the tower from a point on the ground is sin−1\left(

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

9 m

18 m

15 m

12 m

Explanation:

Solution 1 :

Consider a right-angled triangle PQR as shown below.

Let QR = 3 and PR = 5 such that sinθ=$\dfrac{3}{5}$ [i.e., θ=sin⁻¹ $\left( \dfrac{3}{5}\right)$]
PQ =$\sqrt{PR^{2}-QR^{2}}$ (∵ Pythagorean theorem)
i.e., when θ=sin⁻¹$\left(\dfrac{3}{5}\right)$, 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.

Given that θ=sin⁻¹$\left(\dfrac{3}{5}\right)$ and PQ = 20 m
We know that PQ : QR = 4 : 3 (from eq:1)
i.e., 20 : QR = 4 : 3
=> 20 × 3 = QR × 4
=> QR = 15 m
i.e., height of the tower = 15 m

Solution 2 :

Let P be the point of observation and QR be the tower.

Given that θ=sin⁻¹$\left(\dfrac{3}{5}\right)$ and PQ = 20 m
Let the height of the tower, QR = h and PR = x

From the right $\triangle$ PQR,
sinθ=$\dfrac{QR}{PR}$
⇒$sin\left[sin^{−1}\left(\dfrac{3}{5}\right)\right]=\dfrac{h}{x}$
⇒$\dfrac{3}{5}=\dfrac{h}{x}$
⇒x=$\dfrac{5h}{3}$ ...(eq:1)
From Pythagorean theorem, we have
PQ²+QR²=PR²
20²+h²=x²
20²+h²=$\left(\dfrac{5h}{3}\right)^{2}$ (∵ Substituted the value of x from eq:1)
20²+h²=$\left(\dfrac{25h^{2}}{9}\right)$
$\dfrac{25h^{2}}{9}$=20²
$\dfrac{4h}{3}$=20
h=$\dfrac{3×20}{4}$=15 m
i.e., height of the tower = 15 m

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