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If a flagstaff of 6 metres high placed on the top of a tower throws a shadow of \(2 \sqrt{3}\) metres along the ground, then the angle (in degrees) that the sun makes with the ground is
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The correct answer is:
\(60^{\circ}\)
Accordingly,
\(\tan \theta=\frac{h}{x}=\frac{h+6}{x+2 \sqrt{3}}=\frac{6}{2 \sqrt{3}} \Rightarrow \theta=60^{\circ}\)
[Since the triangles \(A E C\) and \(B D C\) are similar] \(h \cot \alpha=(h-100) \cot \beta\)
\(\therefore h=\frac{100 \cot \beta}{\cot \beta-\cot \alpha}\)
\(\tan \theta=\frac{h}{x}=\frac{h+6}{x+2 \sqrt{3}}=\frac{6}{2 \sqrt{3}} \Rightarrow \theta=60^{\circ}\)
[Since the triangles \(A E C\) and \(B D C\) are similar] \(h \cot \alpha=(h-100) \cot \beta\)
\(\therefore h=\frac{100 \cot \beta}{\cot \beta-\cot \alpha}\)
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