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lenght of the shadow of a 15m high pole is 5sqrt3 m at 7o’ clock in the morning . then what is the angle of elevation of sun rays with the ground at the time
Let us take the scenario in the form of a right triangle, where the height of the pole is represented as AB, the shadow of the pole is represented as BC. Let ∠ C be the angle of Elevation.So, We know that,Tan Ф = Opposite / Adjacent=> Tan Ф = 15 / 15 √ 3=> Tan Ф = 1 / √ 3We know that, Tan 30° = 1 / √ 3So we can say that, Ф = 30°Hence the angle of Elevation of the sum at that time was 30°.
Given:height of pole =15mlength of shadow=5√3 RTF: angle of sun’s elevation Solution:Let θ be the angle of sun’s elevationTanθ=15/5√3=√3=tan(π/3)Hence angle of sun’s elevation is π/3
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