Learn all Concepts of Chapter 9 Class 10 (with VIDEOS). Check - Heights and Distances - Class 10

Last updated at Dec. 24, 2019 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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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.