Two towers are situated at a certain distance | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $AB$ be the first tower of height $60 \, m$ and $CD$ be the second tower of height $H \, m$. Let $d$ be the distance between the two towers.Fr

Vedclass · Accepted Answer

$40, 20 \sqrt{3}$

Vedclass · Answer

(A) Let $AB$ be the first tower of height $60 \, m$ and $CD$ be the second tower of height $H \, m$. Let $d$ be the distance between the two towers.From the right-angled triangle $\Delta ABD$,we have:$	an 60^{\circ} = \frac{AB}{BD} = \frac{60}{d}$$\sqrt{3} = \frac{60}{d}$$d = \frac{60}{\sqrt{3}} = 20 \sqrt{3} \, m$.Now,consider the triangle $\Delta ACE$,where $E$ is a point on $AB$ such that $CE \parallel DB$. Then $AE = AB - EB = AB - CD = 60 - H$.In $\Delta ACE$,the angle of depression is $30^{\circ}$,so $\angle ACE = 30^{\circ}$.$	an 30^{\circ} = \frac{AE}{CE} = \frac{60 - H}{d}$$\frac{1}{\sqrt{3}} = \frac{60 - H}{20 \sqrt{3}}$$60 - H = \frac{20 \sqrt{3}}{\sqrt{3}} = 20$$H = 60 - 20 = 40 \, m$.Thus,the height of the second tower is $40 \, m$ and the distance between the towers is $20 \sqrt{3} \, m$.

Similar Questions

If $A$ lies in the third quadrant and $3 \tan A - 4 = 0$,then find the value of $5 \sin 2A + 3 \sin A + 4 \cos A$.

If $\cos A = \frac{3}{4}$,then $32\sin \frac{A}{2}\cos \frac{5A}{2} = $

The value of $\cos 255^o + \sin 195^o$ is

The value of $2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1$ is

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module