A pole 6m high casts slow 2√3m long on the ground then find the sun,s elevation.
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Let AB be the pole and let BC be its shadow at any time.
Now, AB = 6 m and BC = 2√3 m
Let θ be the angular elevation of the sun at that time.
so, ∠ACB = θ
tan θ = perpendicular / base = AB / BC = 6 / 2√3 = √3
⇒ tan θ = √3
Now, AB = 6 m and BC = 2√3 m
Let θ be the angular elevation of the sun at that time.
so, ∠ACB = θ
tan θ = perpendicular / base = AB / BC = 6 / 2√3 = √3
⇒ tan θ = √3
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