# The angle of elev

In the figure, let CD be the tower and AB be the building. Join the points A, D, and B, C. We get two right-angled triangles ∆ABD and ∆BCD, which are right-angled at B and D respectively. We are given that the angle of elevation of the top of the building from the foot of the tower is 30° and that of the top of the tower from the foot of the building is 60°. So, CBD = 60°, ADB = 30°. We are also given that the height of the tower is 60 m. Hence CD = 60 m. We need to find the height of the building, that is AB. For this, we will first find BD from ∆BDC using the trigonometric ratio tan. Using this value of BD, we will find the value of AB from ∆ABD using the trigonometric ratio tan.

In ∆BCD,

or,

So, BD = 60/√3m.

In ∆ABD,

or,

So, height of the building = AB = 20m.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Trigonometric Identities55 mins
Trigonometric Ratios46 mins
Trigonometric Ratios of Specified Angles38 mins
Heights and Distances - I54 mins
Heights and Distances - II45 mins
Reflection of light-238 mins
Purification of Metals60 mins
Extraction of Metal56 mins
Outcomes of Democracy43 mins
Refraction of Light-241 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