KCET · Maths · Heights and Distances
The angle of elevation of the top of a TV tower from three points A, B and \(C\) in a straight line through the foot of the tower are \(\alpha, 2 \alpha\) and \(3 \alpha\)
respectively. If \(A B=a\), then height of the tower is
- A \(a \tan \alpha\)
- B \(a \sin \alpha\)
- C \(a \sin 2 \alpha\)
- D \(a \sin 3 \alpha\)
Answer & Solution
Correct Answer
(C) \(a \sin 2 \alpha\)
Step-by-step Solution
Detailed explanation
In \(\triangle \mathrm{ABE}\),
In \(\triangle \mathrm{BCE}\),
using sine rule,
\(
\begin{aligned}
\frac{\mathrm{a}}{\sin \left(180^{\circ}-3 \alpha\right)} &=\frac{\mathrm{CE}}{\sin 2 \alpha} \\
\Rightarrow \quad \mathrm{CE} &=\frac{\mathrm{a} \sin 2 \alpha}{\sin 3 \alpha}
\end{aligned}
\)
Now, in \(\triangle \mathrm{DCE}\),
\(
\begin{aligned}
&\sin 3 \alpha=\frac{h}{C E} \\
&\Rightarrow \sin 3 \alpha=\frac{h}{a \sin 2 \alpha / \sin 3 \alpha} \quad \text { [from Eq. (i)] } \\
&\Rightarrow \mathrm{h}=\mathrm{a} \sin 2 \alpha
\end{aligned}
\)
In \(\triangle \mathrm{BCE}\),
using sine rule,
\(
\begin{aligned}
\frac{\mathrm{a}}{\sin \left(180^{\circ}-3 \alpha\right)} &=\frac{\mathrm{CE}}{\sin 2 \alpha} \\
\Rightarrow \quad \mathrm{CE} &=\frac{\mathrm{a} \sin 2 \alpha}{\sin 3 \alpha}
\end{aligned}
\)
Now, in \(\triangle \mathrm{DCE}\),
\(
\begin{aligned}
&\sin 3 \alpha=\frac{h}{C E} \\
&\Rightarrow \sin 3 \alpha=\frac{h}{a \sin 2 \alpha / \sin 3 \alpha} \quad \text { [from Eq. (i)] } \\
&\Rightarrow \mathrm{h}=\mathrm{a} \sin 2 \alpha
\end{aligned}
\)
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