A man on the top

The given scene can be represented as a figure shown below

Here AB is the tower with man on it

C is initial position of the car at 30° angle of depression,

D is final position of the car at 45° angle of depression

So, for the car to move from point C to D will take 12 minutes.

We know angle of elevation is equal to angle of depression, so

∠PAC = ∠ACB = 30° and ∠PAD = ∠ADB = 45°……….(i)

And also, the tower is vertical, hence

∠ABC = 90°……….(ii)

Now consider right - angled ΔADB, applying the trigonometric rule,

Now substituting the corresponding values, we get

But tan45° = 1, so above equation becomes,

⇒ AB = BD………..(iii)

Now consider other right - angled ΔACB, applying the trigonometric rule,

Now substituting the corresponding values we get

But , so above equation becomes,

Now comparing equation (iii) and (iv), we get

⇒ √3 BD = BD + DC

⇒ DC = √3 BD - BD

⇒ DC = (√3 - 1) BD

But it is given time taken to cover DC = 12 minutes

So, the time taken to cover minutes

Now we will solve

Multiply and divide by √3 + 1, we get

⇒ 6(√3 + 1) minutes

⇒ 6(1.73 + 1)

= 6(2.73)

= 16.38 minutes

Hence the car will take 16.38 minutes to reach the observation tower from this point.

OR

The given scene can be represented as a figure shown below

Where BE is the water surface

A is 60m above the water surface ⇒ AB = 60m

C is the cloud in the sky

Given the angle of elevation of a cloud from a point 60m above the surface of the water of a lake is 30°, so

∠CAD = 30°

and also given the angle of depression of its shadow from the same point in water of lake is 60°, so

∠DAF = 60°

So, from figure CE will be the height of cloud above the water surface, and EF will be depth of shadow below the water surface.

And AD is a line parallel to water surface BE, hence AD is perpendicular to CF.

So ΔCAD and ΔDAF are right - angled triangle.

Now consider right - angled ΔCAD, applying the trigonometric rule,

Now substituting the corresponding values we get

But , so above equation becomes,

⇒ AD = √3 CD….(i)

Now consider the other right - angled ΔDAF, applying the trigonometric rule,

Now substituting the corresponding values, we get

But tan60° = √3, so above equation becomes,

Equating equation (i) and (ii), we get

⇒ 3CD = DF

But from figure DF = DE + EF, so the above equation becomes,

⇒ 3CD = DE + EF

And we know height of cloud from surface of the water = depth of clouds shadow in the water, so CE = EF, substituting this in above equation we get

⇒ 3CD = DE + CE

But from figure CE = CD + DE, substituting this in above equation, we get

⇒ 3CD = DE + CD + DE

⇒ 3CD - CD = 2DE

⇒ 2CD = 2DE

⇒ CD = DE

Now as AD is parallel to BE, so DE = AB = 60m, substituting this in above equation, we get

⇒ CD = DE = 60m

Hence From figure the height of cloud above the surface of the water is

CE = CD + DE = 60 + 60 = 120m

So, the height of the cloud from the surface of water is 120m.