Question : The angle of elevation of the sun changes from 30° to 45°. The length of the shadow of a pole decreases by 4 metres and the height of the pole is: (Assume $\sqrt3$ = 1.732)
Option 1: 1.464 m
Option 2: 9.464 m
Option 3: 3.648 m
Option 4: 5.464 m
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Correct Answer: 5.464 m
Solution : Let $AB = h$ be the pole. In $\triangle ABC$, $\tan$ 45° = $\frac{AB}{BC}$ ⇒ 1 = $\frac{h}{BC}$ ⇒ $AB = BC = h$ In $\triangle ABD$, $\tan$ 30° = $\frac{h}{h+4}$ ⇒ $\frac{1}{\sqrt3}$ = $\frac{h}{h+4}$ ⇒ $h+4 = \sqrt3h$ ⇒ $h = \frac{4}{\sqrt3-1}$ = $\frac{4(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}$ = $2(\sqrt3+1)$ = 2 × (1.732 + 1) = 5.464 m Hence, the correct answer is 5.464 m.
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