Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.1 , 11 A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see figure). Find the height of the tower and the width of the canal. Let tower be AB So, height of tower = AB Given that,from a point on the other bank directly opposite the Tower,the angle of elevation of the tower is 60°. Hence, ∠ACB = 30° Also, Angle of elevation from point D to top of the tower = 30° So, ∠ ADB = 30° Also, DC = 20 m We need to find height of tower AB & width of canal BC Since tower is vertical to ground So, ∠ ABD = 90° From (1) & (2) √3BC = 𝐵𝐷/√3 √3×√3BC = BD 3BC = BD 3BC = BC + CD 3BC = BC + 20 3BC – BC = 20 2BC = 20 BC = 20/2 BC = 10 m Hence, the width of the canal is 10 m From (1) AB = √3BC AB = √3 × 10 AB = 10√3 m Hence, height of tower = 10√3 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.