Question : The angles of elevation of the top of a temple, from the foot and the top of a building 30 m high, are 60° and 30°, respectively. Then the height of the temple is:
Option 1: 50 metres
Option 2: 43 metres
Option 3: 40 metres
Option 4: 45 metres
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Correct Answer: 45 metres
Solution :
AB = Height of building = 30 metres.
CD = Height of temple = $h$ metres.
AB = CE = 30 metres.
∴ DE = $(h – 30)$ metres;
BC = AE = $x$ metres
$\angle$DAE = 30°; $\angle$DBC = 60°
In ∆BCD,
$\tan 60° = \frac{CD}{BC}$
⇒ $\sqrt{3}=\frac{h}{x}$
⇒ $x=\frac{h}{\sqrt{3}}$---(1)
In ∆ ADE, $\tan 30° = \frac{DE}{AE}$
⇒ $\frac{1}{\sqrt{3}}=\frac{h-30}{x}$
⇒ $x=\sqrt{3}h-30\sqrt{3}$----(2)
Putting the value of $x$ in equation 2 we get,
= $\frac{h}{\sqrt{3}}=\sqrt{3}h – 30\sqrt{3}$
⇒ $3h – h = 30×3$
⇒ $2h = 90$
⇒ $h=45$ metres.
Hence, the correct answer is 45 metres.
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