A man observes the angle of elevation of the | vedclass

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

Question

(B) Let the height of the tower be $CB = 100 \, m$. Let the man walk a distance $AD = x \, m$ towards the tower.In $\Delta DBC$ (right-angled at $B$):

Vedclass · Accepted Answer

$73.2$

Vedclass · Answer

(B) Let the height of the tower be $CB = 100 \, m$. Let the man walk a distance $AD = x \, m$ towards the tower.In $\Delta DBC$ (right-angled at $B$):$	an 45^{\circ} = \frac{CB}{DB}$$1 = \frac{100}{DB}$$DB = 100 \, m$In $\Delta ABC$ (right-angled at $B$):$	an 30^{\circ} = \frac{CB}{AB} = \frac{CB}{AD + DB}$$\frac{1}{\sqrt{3}} = \frac{100}{x + 100}$$x + 100 = 100\sqrt{3}$$x = 100(\sqrt{3} - 1)$Using $\sqrt{3} \approx 1.732$:$x = 100(1.732 - 1) = 100(0.732) = 73.2 \, m$.Thus,the distance walked by the man is $73.2 \, m$.

$A$ man observes the angle of elevation of the top of a tower having a height of $100 \, m$ as $30^{\circ}$. After walking a certain distance towards the tower,he observes the angle of elevation as $45^{\circ}$. What is the distance through which the man walked towards the tower? (in $m$)

Similar Questions

$\sin 15^\circ + \cos 105^\circ = $

If $\tan (A - B) = 1$ and $\sec (A + B) = \frac{2}{\sqrt{3}}$,then the smallest positive value of $B$ is

The value of $\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ$ is

From an aeroplane flying vertically over a straight horizontal road,the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to be $30^{\circ}$ and $60^{\circ}$. Find the height of the aeroplane above the road in miles.

If $\sec \alpha + \tan \alpha = 2,$ then the value of $\sin \alpha$ is (assume that $0 < \alpha < 90^{\circ}$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module