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A pole 6 m high casts a shadow 2\(\sqrt 3 \)m long on the ground, then the sun’s elevation is
  • \(45^\circ \)
  • \(60^\circ \)
  • \(75^\circ \)
  • \(30^\circ \)
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Question

A pole 6 m high casts a shadow 2\(\sqrt 3 \)m long on the ground, then the sun’s elevation is
  • \(45^\circ \)
  • \(60^\circ \)
  • \(75^\circ \)
  • \(30^\circ \)

Answer

Let height = 6m
length of shadow = \(2 \sqrt{3 m}\)
\(\theta\) is angle of elevation
tan\(\theta\) = (height) / (shadow length)
\( = \frac{6}{2}\sqrt 3 = \sqrt 3 \)
\(\theta = \frac{\pi }{3}\)
Angle of inclination is = 60o
Therefore the choice is: B
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