If a pole 6 m high casts a shadow 2√3m long on the ground, then the Sun’s elevation is

(a) 60°                      (b) 45°                     (c) 30°                  (d) 90°

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Given Height of pole = DE = 6 m Length of shadow = EF = 2βˆšπŸ‘m We need to find Angle of elevation of the sun i.e 𝛉 In βˆ† 𝐷𝐸𝐹 tan πœƒ = (𝑆𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’" " 𝐹)/β–ˆ(𝑆𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’" " 𝐹@ ) tan 𝜽 = (" " 𝑫𝑬)/𝑬𝑭 tan πœƒ = (" " 6)/(2√3) tan πœƒ = (" " 3)/√3 tan πœƒ =√3 ∴ 𝜽 = 60 Β° So, the correct answer is (a)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo