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
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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