# Question 1 - Questions easy to difficult - Chapter 9 Class 10 Some Applications of Trigonometry

Last updated at April 16, 2024 by Teachoo

Questions easy to difficult

Angle of Elevation and Angle of Depression

Angle of Depression from point A to point B is same as Angle of Elevation from point B to point A

How to find Height when angle of elevation is given?

How to find Distance when angle of depression is given?

Example 1

EX 9.1, 4

EX 9.1, 5

EX 9.1, 3 Important

EX 9.1, 2 Important

Ex 9.1, 7

Example 4

Example 2 Important

EX 9.1, 8

Example 5

Ex 9.1, 11

Example 7 Important

Example 3 Important

Ex 9.1, 6 Important

EX 9.1, 9 Important

Ex 9.1, 10

Example 6 Important

Ex 9.1, 13 Important

Ex 9.1, 12 Important

Question 1 Important Deleted for CBSE Board 2025 Exams You are here

Ex 9.1, 14 Important

Ex 9.1, 15 Important

EX 9.1, 1

Last updated at April 16, 2024 by Teachoo

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 𝜶 Angle of elevation from Q is 𝜷. Given 𝛼 and 𝛽 are complementary. 𝜶 + 𝜷 = 90° We need to prove AB = 6 m In ∆ABP tan 𝜶 = 𝑨𝑩/𝑩𝑷 In ∆ABQ tan 𝜷 = AB/BQ tan (90 − 𝜶) = AB/BQ cot 𝜶 = AB/BQ 1/(tan α)=AB/BQ tan 𝛂 = 𝑩𝑸/𝐀𝐁 From (1) and (2) 𝑨𝑩/𝑩𝑷 = 𝑩𝑸/𝑨𝑩 AB × AB = BQ × BP AB2 = 4 × 9 AB2 = 36 AB = (36)^(1/2) AB = 6 m Hence, height of the tower is 6m. Hence proved