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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