From a point on the ground, the angle of elevation of the top of a vertical tower is found to be such that its tangent is $\frac{3}{5}$ . On walking $50 m$ towards the tower, the tangent of the new angle of elevation of the top of the tower is found to be $\frac{4}{5}$ . Find the height of the tower.

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$121 m$

$150 m$

$120 m$

$160 m$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Let $A B$ is the tower and $A B = h$

$P, Q$ are the observation points

Given $P Q = 50 m$

$\tan α = \frac{3}{5}, \tan β = \frac{4}{5}$

$\begin{matrix} In Δ A B P & In Δ A B Q \\ \tan α = \frac{A B}{P B} & \tan β = \frac{A B}{B Q} \\ \frac{3}{5} = \frac{h}{50 + x} & \frac{4}{5} = \frac{h}{x} \\ 5 h = 150 + 3 x \to (1) & 5 h = 4 x \to (2) \end{matrix}$

From (1) and (2)

$4 x = 150 + 3 x$

$x = 150 m$

Sub $x = 150$ in (1)

$5 h = 150 + 450$

$5 h = 600$

$h = 120 m$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Get Expert Academic Guidance – Connect with a Counselor Today!