Question : The shadow of a tower when the angle of elevation of the sun is 45°, is found to be 10 metres longer than when it was 60°. The height of the tower is:
Option 1: $5\sqrt{3}-1$ metres
Option 2: $5(3+\sqrt{3})$ metres
Option 3: $10(\sqrt{3}-1)$ metres
Option 4: $10(\sqrt{3}+1)$ metres
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Correct Answer: $5(3+\sqrt{3})$ metres
Solution : Here, AB is the tower. BD is the shadow when $\angle$ADB = 45°, CB is the shadow when $\angle$ACB = 60°. By the given condition, CD = 10 m. In $\triangle$ADB, tan45° = $\frac{AB}{BD}$ ⇒$\frac{AB}{BC+CD} =1$ ⇒ AB = BC + 10 --(1) In △ACB, $\tan60° = \frac{AB}{BC}$ ⇒ $AB =\sqrt{3}BC$ --(2) From 1 and 2 we have, $BC + 10 = \sqrt{3}BC$ ⇒ $BC =\frac{10}{\sqrt{3}-1}=5(\sqrt{3}+1)$ m. $∴ AB = BC + 10 =5(\sqrt{3}+1)+10=5(\sqrt{3}+3)$ m Hence, the correct answer is $5(3+\sqrt{3})$ m.
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