Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.1 , 7 From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. Let the building be AB and tower be CA. Building is 20 m high, So, AB = 20m Let point of ground be P. Angle of elevation from point P to bottom of tower A = 45° Hence, ∠APB = 45° Also, Angle of elevation from point P to top of the tower C = 60° Hence, ∠CPB = 60° We need to find AC. Since building is vertical, ∠ ABP = 90° . In right angle triangle APB, tan P = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃) tan P = 𝐴𝐵/𝑃𝐵 tan 45° = 20/𝑃𝐵 1 = 20/𝑃𝐵 PB = 20 m Similarly, in right angle triangle CPB, tan P = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃) tan P= 𝐵𝐶/𝑃𝐵 tan 60° = 𝐵𝐶/𝑃𝐵 √3 = 𝐵𝐶/20 20√3 = BC BC = 20√3 AB + AC = 20√3 20 + AC = 20√3 AC = 20√3 – 20 AC = 20 (√3 – 1) m Hence, Height of tower = AC = 20 (√3 – 1) m

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.