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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