Question

(1) The angle of elevation of a cloud from a point 60 m above a lake is 30° and the
angle of depression of the reflection of the cloud in the lake is 60°. Find the height
of the cloud from the surface of the lake.​

Answer 1

Let AO = H

CD = OB = 60m

A'B = AB = (60+H) m

In triangle AOD,

tan30° = AO/OD = H/OD

H=OD/√3

OD=√3 H

Now, in triangle A'OD,

tan60° = OA'/OD = (OB+BA')/OD

√3 = (60+60+H)/√3 H

    =(120+H)/√3 H

 => 120+H = 3H

     2H = 120

     H = 60 m

Thus, height of the cloud above the lake = AB+A'B

                                                                    =(60+60)

                                                                    = 120m

Hope it helps..

Answer 2

$\huge\mathfrak\pink{Solution :-}$

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 =

$\rm Draw \: PM \perp CBDrawPM⊥CB$

Let CM = h

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

$\rm In \: \triangle \: CPMIn△CPM$

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

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

$\rm ➨ PM = \sqrt{3} h.......(i)➨PM=$

$\rm In \: \triangle \: PMC,In△PMC,$

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

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

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

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

$\rm☄from \: eq(i) \: and \: eq(ii)$

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

$\rm \implies3h = h + 120m⟹3h=h+120m$

$\rm \implies3h - h= 120m⟹3h−h=120m$

$\rm \implies2h = 120⟹2h=120$

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

$\rm \implies h = 60m⟹h=60m$

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

$⇾ \rm Thus,the \: height \: of \: the \: cloud$

$\: from \: the \: surface \:of \: lake \boxed{ \rm120m.}$