Q. 14.6( 20 Votes )

A tower stands vertically on the ground. From a point on the ground which is 20 m away from the foot of the tower, the angle of elevation of its top is found to be 60°. Find the height of the tower. [Take √3 = 1.732.]

Answer :


Let AB be the tower and C be the point on the ground 20 m away from the foot of the tower B from where the angle of elevation of the top of the tower is 60°. Now, draw a line from C to A. Join B and C. We get a triangle ABC with right angle at B. We are to find the height of the tower, that is AB. We will be using trigonometric ratios involving AB(height) and BC(base). So, ACB = 60° BC = 20m.

Now in ∆ABC,


AB = 20√3 = 34.64m.

The height of the tower is 34.64 m.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Heights and Distances33 mins
Quiz | Imp. Qs. on Heights and Distances37 mins
NCERT | Basics of Heights and Distances37 mins
Idioms and Phrases43 mins
Goprep Genius Quiz | Analogy and Classification48 mins
Acid - Types and Nomenclature52 mins
Agriculture and its Importance35 mins
Foundation | Permutation and Combination45 mins
Consistent and Inconsistent Equations33 mins
Verb & Tenses46 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses