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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