Question

The angle of elevation of the cloud from a point 60 m above take is 30° and the

angle of depression of the reflection of the cloud in the take is 60°. Find the

height of the cloud.​

Answer 1

CORRECT QUESTION.

The angle of a elevation of a cloud from a point

60 m above the surface to the water of a lake

30° and the angle of depression of its shadow

from the same point in water lake is 45°

find the height of the cloud from the surface

of the water.

EXPLANATION.

$\sf : \implies \: according \: to \: the \: question \\ \\ \sf : \implies \: in \: right \: angled \: triangle \: = \triangle \: bec \\ \\ \sf : \implies \: \tan( \theta) = \frac{perpendicular}{base} \\ \\ \sf : \implies \: \tan(30 \degree) = \frac{be}{bc} \\ \\ \sf : \implies \: \frac{1}{ \sqrt{3} } = \frac{h - 60}{be} \\ \\ \sf : \implies \: be \: = \sqrt{3}(h - 60) \: \: .....(1)$

$\sf : \implies \: in \: right \: angle \: triangle \: = \triangle \: bed \\ \\ \sf : \implies \: \tan( \theta) = \frac{perpendicular}{base} \\ \\\sf : \implies \: \tan(60 \degree) = \frac{h + 60}{be} \\ \\ \sf : \implies \: \sqrt{3} = \frac{h + 60}{be} \\ \\ \sf : \implies \: \sqrt{3}be = h + 60 \: \: ....(2)$

$\sf : \implies \: from \: equation \: (1) \: \: and \: \: (2) \\ \\ \sf : \implies \: put \: the \: value \: of \: equation \: (1) \: in \: (2) \\ \\ \sf : \implies \: \sqrt{3} \times \sqrt{3} (h - 60) = h + 60 \\ \\ \sf : \implies \: 3(h - 60) = h + 60 \\ \\ \sf : \implies \: 3h - 180 = h + 60 \\ \\ \sf : \implies \: 2h = 240 \\ \\ \sf : \implies \: h = 120 \: m \\ \\ \sf : \implies \: \green{{ \underline{height \: of \: cloud \: = 120 \: m \:}}}$

Answer 2

Let AB be the surface of the lake and P be the point of observation such that AP = 60 m.

Let C be the position of the cloud and C be its reflection in the lake.

Then CB =

$\rm Draw \: PM \perp CB$

Let CM = h

$\therefore \rm CB = h + 60 m$

$\rm In \: \triangle \: CPM$

$\rm \tan30 \degree = \frac{CM}{PM}$

$\rm ➨ \frac{1}{ \sqrt{3} } = \frac{h}{PM}$

$\rm ➨ PM = \sqrt{3} h.......(i)$

$\rm In \: \triangle \: PMC,$

$\rm ➨\tan60 \degree = \frac{C ' M}{PM}$

$\rm ➨ \tan60 \degree = \frac{C'B + BM}{PM}$

$\rm ➨ \sqrt{3} = \frac{h + 60m + 60m}{PM}$

$\rm ➨ \sqrt{3} = \frac{h +120m}{PM} .......(ii)$

$\rm☄from \: eq(i) \: and \: eq(ii)$

$\rm ➨ \sqrt{3}h = \frac{h + 120m}{ \sqrt{3} }$

$\rm \implies3h = h + 120m$

$\rm \implies3h - h= 120m$

$\rm \implies2h = 120$

$\rm \implies h = \frac{ \cancel{120}}{ \cancel{2} } = 60m$

$\rm \implies h = 60m$

$\rm ➯ CB = h + 60m = 60m + 60m = 120m$

$⇾ \rm Thus,the \: height \: of \: the \: cloud \: from \: the \: surface \\ \rm of \: lake \boxed{ \rm120m.}$