A person walking along a straight road toward | vedclass

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

Question

(A) Let $h$ be the height of the hill $QP$ and $x$ be the distance from the second point $A$ to the base $P$ of the hill.In $	riangle QPA$,$	an 60^{

Vedclass · Accepted Answer

$\frac{3}{2} 	ext{ km}$

Vedclass · Answer

(A) Let $h$ be the height of the hill $QP$ and $x$ be the distance from the second point $A$ to the base $P$ of the hill.In $	riangle QPA$,$	an 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3}x$.In $	riangle QPO$,the distance $OA = \sqrt{3} 	ext{ km}$,so $OP = OA + AP = \sqrt{3} + x$.$	an 30^{\circ} = \frac{h}{\sqrt{3} + x} \implies \frac{1}{\sqrt{3}} = \frac{h}{\sqrt{3} + x} \implies \sqrt{3}h = \sqrt{3} + x$.Substitute $h = \sqrt{3}x$ into the equation: $\sqrt{3}(\sqrt{3}x) = \sqrt{3} + x \implies 3x = \sqrt{3} + x \implies 2x = \sqrt{3} \implies x = \frac{\sqrt{3}}{2} 	ext{ km}$.Therefore,$h = \sqrt{3} 	imes \frac{\sqrt{3}}{2} = \frac{3}{2} 	ext{ km}$.

$A$ person walking along a straight road towards a hill observes at two points,separated by a distance of $\sqrt{3} \text{ km}$,that the angles of elevation of the hill are $30^{\circ}$ and $60^{\circ}$. The height of the hill is

Similar Questions

From the top of a $25 \ m$ high cliff,the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find the height of the tower in $m$.

The angle of elevation of the top of a tower from a point $20 \ m$ away from its base is $45^{\circ}$. The height of the tower is (in $m$):

An observer on the top of a cliff,$200 \, m$ above the sea level,observes the angle of depression of two ships at anchor to be $45^{\circ}$ and $30^{\circ}$ respectively. Find the distance (in $m$) between the ships if the line joining them stretches to the base of the cliff.

The height of a tower is $100 \text{ m}$. When the angle of elevation of the sun changes from $30^{\circ}$ to $45^{\circ}$,the shadow of the tower becomes $x \text{ m}$ smaller. The value of $x$ is (in $\text{m}$):

The ratio of the length of a rod and its shadow is $1 : \sqrt{3}$. The angle of elevation of the sun is (in $^{\circ}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module