Question : A 1.6 m tall observer is 45 metres away from a tower. The angle of elevation from his eye to the top of the tower is 30°, then the height of the tower in metres is: (Take$\sqrt{3}=1.732$)
Option 1: 25.98
Option 2: 26.58
Option 3: 27.58
Option 4: 27.98
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Correct Answer: 27.58
Solution :
AB = Height of observer = 1.6 m.
CD = Height of tower = $h$ m.
∴ DE = ($h$ – 1.6) m
BC = AE = 45 metres
$\angle$DAE = 30°
From $\triangle$DAE,
$\tan 30°=\frac{DE}{AE}$
⇒ $\frac{1}{\sqrt{3}}=\frac{h-1.6}{45}$
⇒ $h-1.6=15\sqrt{3}$
⇒ $h – 1.6 = 15 × 1.732$
⇒ $h = (25.98 + 1.6)$
$\therefore h=27.58$ metres
Hence, the correct answer is 27.58 metres.
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