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Ex 9.1, 4 - The angle of elevation of top of a tower - Questions easy to difficult

EX 9.1, 4 - Part 2

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Ex 9.1 , 4 The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower. Let tower be AB Let point be C Distance of point C from foot of tower = 30m Hence, BC = 30m Angle of elevation = 30° So, ∠ACB = 30° Since tower is vertical, ∠ ABC = 90° We need to find height of tower i.e. AB In right triangle ABC tan C = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" c " )/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝐶) tan 30° = (" " AB)/BC 1/√3 = (" " AB)/30 30/√3 = AB AB = 30/√3 Multiplying √3 in numerator and denominator AB= 30/√3 × √3/√3 AB = (30√3)/3 AB = 10√3 Hence, Height of tower = AB = 10√3 m

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.