4 Views

Question : From the top of a 20 metres high building, the angle of elevation from the top of a tower is 60° and the angle of depression of its foot is at 45°, then the height of the tower is: $(\sqrt{3} = 1.732)$

Option 1: 45.46 metres

Option 2: 45.64 metres

Option 3: 54.64 metres

Option 4: 54.46 metres


Team Careers360 10th Jan, 2024
Answer (1)
Team Careers360 22nd Jan, 2024

Correct Answer: 54.64 metres


Solution :
In $\triangle BDE$,
$\tan 45°=\frac{BD}{DE}$
⇒ $1=\frac{20}{DE}$
⇒ $DE=20$ metres
⇒ $DE=BC=20$ metres
In $\triangle ABC$,
$\tan 60°=\frac{AC}{BC}$
⇒ $\sqrt{3}=\frac{AC}{20}$
⇒ $AC=20\sqrt{3}=20\times1.732=34.64$ metres
So, the height of the tower is (34.64 + 20) = 54.64 metres
Hence, the correct answer is 54.64 metres.

SSC CGL Complete Guide

Candidates can download this ebook to know all about SSC CGL.

Download EBook

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books