Question : If the elevation of the Sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 metres high, is:
Option 1: $7.5$ metres
Option 2: $15$ metres
Option 3: $10\sqrt{3}$ metres
Option 4: $5\sqrt{3}$ metres
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Correct Answer: $10\sqrt{3}$ metres
Solution : Let AB be the pole. Suppose BD and BC are the lengths of the shadow of the pole when the elevation of the Sun is 30° and 60°, respectively. Here, AB = 15 metres In right $\triangle$ABC, $⇒\tan 60°=\frac{\text{AB}}{\text{BC}}$ $⇒\sqrt{3}=\frac{AB}{\text{BC}}$ $⇒BC= \frac{15}{\sqrt3} =5\sqrt3$ metres .................... (i) In $\triangle$ABD, $⇒\tan 30°=\frac{\text{AB}}{\text{BD}}$ $⇒\frac{1}{\sqrt{3}}=\frac{15}{\text{$5\sqrt3$+CD}}$ $⇒5\sqrt3 + CD = 15\sqrt3$ $⇒CD = 10\sqrt3$ metres Now, if the elevation of the sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 m high is $10\sqrt{3}$ metres. Hence, the correct answer is $10\sqrt{3}$ metres.
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