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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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.