From the top of a vertical tower, the angles of depression of two cars in the same straight line with the base of the tower, at an instant are found to be 45° and 60°. If the cars are 100 m apart and are on the same side of the tower, find the height of the tower.

In the above figure, let AB be the tower and P and D be the positions of the two cars at an instant, observed from A. join A and E. The angles of depression are DAE and PAE. Given that, PD = 100 m. Since AE is parallel to BD, so, ADB = DAE = 45° and APB = PAE = 60°. Join P,D and A,B. We get two right-angled triangles ∆ABD and ∆ABP. We are to find AB. We use trigonometric ratio tan for both the triangles, using AB as height and BP as base for ∆ABP and AB as height and BD as base for ∆ABD.

From ∆ABD,

or, AB = BD

From ∆APB,

or, BD = BP√3

or, BP + 100 = BP√3

or, BP(√3-1) = 100

or,

Hence, the height of the tower is, AB = BD = BP + 100 = 136.61 + 100 = 236.61m

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
Learn to Make Your Own Acid Rain and pH Paper at Home!27 mins
Class and Object in JAVA61 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