From an aeroplane flying vertically over a st | vedclass

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

Question

(B) Let $P$ be the position of the aeroplane and $S$ be the point on the road directly below $P$. Let $PS = h$ be the height of the aeroplane.Let $Q$

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{4}$

Vedclass · Answer

(B) Let $P$ be the position of the aeroplane and $S$ be the point on the road directly below $P$. Let $PS = h$ be the height of the aeroplane.Let $Q$ and $R$ be the positions of the two consecutive milestones on the road. Since they are consecutive milestones,the distance between them is $QR = 1$ mile.Given that the angles of depression are $30^{\circ}$ and $60^{\circ}$,the angles of elevation from the milestones to the aeroplane are also $30^{\circ}$ and $60^{\circ}$ respectively.In $\Delta PQS$,$	an 30^{\circ} = \frac{PS}{QS} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{QS} \Rightarrow QS = h\sqrt{3}$.In $\Delta PSR$,$	an 60^{\circ} = \frac{PS}{SR} \Rightarrow \sqrt{3} = \frac{h}{SR} \Rightarrow SR = \frac{h}{\sqrt{3}}$.Since $QS + SR = QR = 1$ mile,we have $h\sqrt{3} + \frac{h}{\sqrt{3}} = 1$.Multiplying by $\sqrt{3}$,we get $3h + h = \sqrt{3} \Rightarrow 4h = \sqrt{3} \Rightarrow h = \frac{\sqrt{3}}{4}$ miles.

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.

Similar Questions

$\text{cosec } A - 2 \cot 2A \cos A = $

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$ is equal to

If $\tan \alpha = \frac{m}{m+1}$ and $\tan \beta = \frac{1}{2m+1}$,then $\alpha + \beta =$

$\log _{\frac{1}{8}\csc^2 \frac{\pi}{8}} \sin^2 \frac{3\pi}{8}$ equals to

The minimum value of $2 \sin ^{2} \theta+3 \cos ^{2} \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module