Example 5 - Chapter 9 Class 10 Some Applications of Trigonometry

Last updated at Dec. 13, 2024 by

Example 5 - The shadow of a tower standing on a level ground - Examples

Example 5 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 2
Example 5 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 3

Go Ad-free

Transcript

Example 5 The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. Given tower be AB When Sun’s altitude is 60° ∠ ACB = 60° & Length of shadow = BC When Sun’s altitude is 30° ∠ ADB = 30° & Length of shadow = DB Shadow is 40 m when angle changes from 60° to 30° CD = 40 m We need to find height of tower i.e. AB Since tower is vertical to ground ∴ ∠ ABC = 90° From (1) & (2) 𝐴𝐵/√3 = √3 AB – 40 AB" =" √3(√3 " AB) – 40" √3 AB = 3AB –" 40" √3 "40" √3 = 3AB – AB "40" √3 = 2AB 2AB = "40" √3 AB = (40√3)/2 AB = 20√3 Hence, Height of the tower = AB = 20√(3 )metre

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo