Question : From point P on the level ground, the angle of elevation to the top of the tower is 30°. If the tower is 100 metres high, the distance of point P from the foot of the tower is: (Take $\sqrt{3}$ = 1.73)
Option 1: 149 m
Option 2: 156 m
Option 3: 173 m
Option 4: 188 m
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Correct Answer: 173 m
Solution :
Let AB be the tower.
Given: AB = 100 m and $\theta$ = 30°
In $\triangle$ PAB,
$\tan$ $\theta$ = $\frac{AB}{PB}$
⇒ $\tan 30°$ = $\frac{100}{PB}$
⇒ $\frac{1}{\sqrt{3}}$ = $\frac{100}{PB}$
⇒ $PB = 100 × \sqrt{3} = 100 × 1.73 = 173$ m
Hence, the correct answer is 173 m.
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