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Question 1 The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m. Given that AB is the tower P, Q are the point at distance 4 m and 9 m resp. Also, PB = 4 m , QB = 9 m & Angle of elevation from P is ๐œถ Angle of elevation from Q is ๐œท. Given ๐›ผ and ๐›ฝ are complementary. ๐œถ + ๐œท = 90ยฐ We need to prove AB = 6 m In โˆ†ABP tan ๐œถ = ๐‘จ๐‘ฉ/๐‘ฉ๐‘ท In โˆ†ABQ tan ๐œท = AB/BQ tan (90 โˆ’ ๐œถ) = AB/BQ cot ๐œถ = AB/BQ 1/(tan ฮฑ)=AB/BQ tan ๐›‚ = ๐‘ฉ๐‘ธ/๐€๐ From (1) and (2) ๐‘จ๐‘ฉ/๐‘ฉ๐‘ท = ๐‘ฉ๐‘ธ/๐‘จ๐‘ฉ AB ร— AB = BQ ร— BP AB2 = 4 ร— 9 AB2 = 36 AB = (36)^(1/2) AB = 6 m Hence, height of the tower is 6m. Hence proved

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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.