Question : The shadow of a tower standing on a level plane is 40 metres longer when the sun's altitude is 45°, than when it is 60°. The height of the tower is:
Option 1: $30(3+\sqrt{3})$ metres
Option 2: $40(3+\sqrt{3})$ metres
Option 3: $20(3+\sqrt{3})$ metres
Option 4: $10(3+\sqrt{3})$ metres
Correct Answer: $20(3+\sqrt{3})$ metres
Solution :
Given: Height of the tower = AB
Shadow at 60° = BC
Shadow at 45° = BD
CD = 40 metres
Let BC be $x$ m.
In $\Delta$ ABC, we have,
$\tan 60°=\frac{AB}{BC}$
⇒ $\sqrt{3}=\frac{AB}{x}$
⇒ $x=\frac{AB}{\sqrt{3}}$ -------(1)
In $\Delta$ ABD, we have,
$\tan 45°=\frac{AB}{BD}$
⇒ $1=\frac{AB}{x+40}$
⇒ $x+40=AB$
⇒ $x=AB-40$ ------(2)
From equations (1) and (2), we have,
⇒ $\frac{AB}{\sqrt{3}}=AB-40$
⇒ $AB=\sqrt3AB-40\sqrt{3}$
⇒ $AB=\frac{40\sqrt{3}}{\sqrt3-1}$
⇒ $AB=\frac{40\sqrt3(\sqrt3+1)}{2}$
⇒ $AB=20(3+\sqrt{3})$ metres
Hence, the correct answer is $20(3+\sqrt{3})$ metres.
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