Q. 274.2( 4 Votes )

A man on the top

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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