Answer :

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 is:

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 the water of the lake is 60°, so

∠DAF = 60°

So, from figure CE will be the height of cloud above the water surface, and EF will be the 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 the 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 the height of cloud from surface of the water = depth of clouds shadow in the water,

so CE = EF, substituting this in the above equation we get,

⇒ 3CD = DE + CE

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

⇒ CD = DE = 60m

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

CE = CD + DE

= 60 + 60

= 120m

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

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