# Ex 9.1, 16 - Chapter 9 Class 10 Some Applications of Trigonometry

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.