A pole 6 m high casts a shadow 2\(\sqrt 3 \)m long on the ground, then the sun’s elevation is

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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

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 = 60^{o}

Therefore the choice is: B

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 = 60

Therefore the choice is: B

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