From the top of a building …

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Ricky Roy 6 years, 11 months ago
- 4 answers
Gaurav Seth 6 years, 11 months ago
Let BC be the building and AD be the tower.
Let the height of tower, AD be h m.
Angles of depression of the top D and the bottom A of the tower CB are 30° and 60° respectively.
∴ ∠CDE = 30°
∠CAB = 60°
Since, BC = 60 m.
∴ CE = (60 – h) m
Let AB = DE = x m
In ∆DEC,

In ∆CBA,

Equating equation (1) and (2),

⇒ 3 (60 – h) = 60
⇒ 180 – 3h) = 60
⇒ 180 – 60 = 3n
⇒ 120 = 3h

⇒ h = 40
Thus, the height of the tower is 40 m.
Related Questions
Posted by Vanshika Bhatnagar 1 year, 4 months ago
- 2 answers
Posted by Lakshay Kumar 1 year, 1 month ago
- 0 answers
Posted by Hari Anand 6 months, 1 week ago
- 0 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 0 answers
Posted by Kanika . 1 month, 1 week ago
- 1 answers
Posted by Sahil Sahil 1 year, 4 months ago
- 2 answers
Posted by Parinith Gowda Ms 3 months, 3 weeks ago
- 1 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Prince Garg 6 years, 11 months ago
0Thank You