# Ex 9.1, 16

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.1, 16 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 4m and 9m resp. Also, PB = 4m , QB = 9m & Angle of elevation from P is 𝛼 Angle of elevation from Q is 𝛽. Given 𝛼 and 𝛽 are supplementary. 𝛼 + 𝛽 = 90° We need to prove AB = 6m From (1) and (2) ABBP = BQAB AB × AB = BQ × BP (AB)2= 4 × 9 (AB)2= 36 AB = 36 12 AB = 6m Hence, height of the tower is 6m. Hence proved

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .