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Question: The shadow of a tower standing on a level ground is found to be 60 m longer when the sun's altitude ...

The shadow of a tower standing on a level ground is found to be 60 m longer when the sun's altitude is 30∘30 ^ { \circ } than when it is 45∘45 ^ { \circ }. The height of the tower is

A

60 m

B

30 m

C

603m60 \sqrt { 3 } m

D

30(3+1)m30 ( \sqrt { 3 } + 1 ) m

Answer

30(3+1)m30 ( \sqrt { 3 } + 1 ) m

Explanation

Solution

∙∙\bullet \bullet AB=AM−BMA B = A M - B M ⇒ ABh=AMh−BMh\frac { A B } { h } = \frac { A M } { h } - \frac { B M } { h }

ABh=cot⁡30∘−cot⁡45∘\frac { A B } { h } = \cot 30 ^ { \circ } - \cot 45 ^ { \circ }

⇒ h=603−1=60(3+1)(3−1)(3+1)h = \frac { 60 } { \sqrt { 3 } - 1 } = \frac { 60 ( \sqrt { 3 } + 1 ) } { ( \sqrt { 3 } - 1 ) ( \sqrt { 3 } + 1 ) }

⇒h=60(3+1)3−1h = \frac { 60 ( \sqrt { 3 } + 1 ) } { 3 - 1 }

⇒ h=30(3+1)mh = 30 ( \sqrt { 3 } + 1 ) m