The angle of elevation of a cloud C from a po | vedclass

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

Question

(A) Let $PA = x$ be the horizontal distance from $P$ to the vertical line passing through the cloud $C$.Let $A$ be the point on the vertical line such

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(A) Let $PA = x$ be the horizontal distance from $P$ to the vertical line passing through the cloud $C$.Let $A$ be the point on the vertical line such that $PA \perp AC$.In $\Delta PAC$,$	an(30^{\circ}) = \frac{AC}{PA}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{AC}{x}$ $\Rightarrow AC = \frac{x}{\sqrt{3}}$.Let $B$ be the point on the surface of the lake directly below $P$. Then $PB = 200 \ m$.The height of the cloud above the lake is $H = AC + AB = AC + 200 = \frac{x}{\sqrt{3}} + 200$.The image of the cloud $C'$ is at a depth $H$ below the surface of the lake.So,$BC' = H = \frac{x}{\sqrt{3}} + 200$.In $\Delta PBC'$,the total vertical distance is $AC' = AB + BC' = 200 + (\frac{x}{\sqrt{3}} + 200) = 400 + \frac{x}{\sqrt{3}}$.The angle of depression is $60^{\circ}$,so $	an(60^{\circ}) = \frac{AC'}{PA} = \frac{400 + x/\sqrt{3}}{x}$.$\sqrt{3} = \frac{400 + x/\sqrt{3}}{x} \Rightarrow \sqrt{3}x = 400 + \frac{x}{\sqrt{3}}$.Multiply by $\sqrt{3}$: $3x = 400\sqrt{3} + x$ $\Rightarrow 2x = 400\sqrt{3}$ $\Rightarrow x = 200\sqrt{3}$.In $\Delta PAC$,$PC = \frac{PA}{\cos(30^{\circ})} = \frac{x}{\sqrt{3}/2} = \frac{2x}{\sqrt{3}} = \frac{2(200\sqrt{3})}{\sqrt{3}} = 400 \ m$.

The angle of elevation of a cloud $C$ from a point $P$,$200 \ m$ above a still lake is $30^{\circ}$. If the angle of depression of the image of $C$ in the lake from the point $P$ is $60^{\circ}$,then $PC$ (in $m$) is equal to

Similar Questions

$A$ $20 \ m$ long tree is broken by the wind such that its top touches the ground at an angle of $30^\circ$. The height from the ground at which the tree is broken is:

The shadow of a tower is found to be $60 \ m$ shorter when the sun's altitude changes from $30^{\circ}$ to $60^{\circ}$. The height of the tower from the ground is approximately equal to......$m$

For a man,the angle of elevation of the highest point of the temple situated east of him is $60^\circ$. On walking $240 \ m$ to the north,the angle of elevation is reduced to $30^\circ$. The height of the temple is:

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object, the angle of elevation is found to be $\beta$. Then the height (in metres) of the object is

The angle of elevation of an object on a hill is observed from a certain point in the horizontal plane through its base to be $30^{\circ}$. After walking $120 \ m$ towards it on level ground, the angle of elevation is found to be $60^{\circ}$. Then the height of the object (in metres) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module