# Ex 9.1, 16 - Chapter 9 Class 10 Some Applications of Trigonometry (Term 2)

Last updated at Jan. 31, 2022 by Teachoo

Chapter 9 Class 10 Some Applications of Trigonometry (Term 2)

Serial order wise

Last updated at Jan. 31, 2022 by Teachoo

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 4 m and 9 m resp. Also, PB = 4 m , QB = 9 m & Angle of elevation from P is πΆ Angle of elevation from Q is π·. Given πΌ and π½ are complementary. πΆ + π· = 90Β° We need to prove AB = 6 m In βABP tan πΆ = π¨π©/π©π· In βABQ tan π· = AB/BQ tan (90 β πΆ) = AB/BQ cot πΆ = AB/BQ 1/(tan Ξ±)=AB/BQ tan π = π©πΈ/ππ From (1) and (2) π¨π©/π©π· = π©πΈ/π¨π© AB Γ AB = BQ Γ BP AB2 = 4 Γ 9 AB2 = 36 AB = (36)^(1/2) AB = 6 m Hence, height of the tower is 6m. Hence proved