Question

A man from top of a 100 meters high tower sees a car moving towards the tower at an angle of depression of $ 30^\circ $ . After some time, the angle of depression becomes $ 60^\circ $ . The distance (In meters) travelled by the car during this time is
A) \[100\sqrt 3 \]
B) $ \dfrac{{200\sqrt 3 }}{3} $
C) $ \dfrac{{100\sqrt 3 }}{3} $
D) $ 200\sqrt 3 $

Answer 1

Hint: To solve the above problem we have to put the data in a diagram of the right angled triangle. From that first we will derive the distance of the tower from the final point by taking a tangent of angle of depression at the final point. Then using the distance derived in the equation of tangent of depression angle at the starting point we will get the distance travelled by the car.

Complete step-by-step answer:
According to the question, the man is on the top of a 100 meters high tower.
Hence, the length of the tower is 100 meters.
The angle of depression first was $ 30^\circ $ , then it became $ 60^\circ $ .
So, let $ {\theta _1} = 30^\circ $ and $ {\theta _2} = 60^\circ $
Representing the above data in the below diagram,

From the above figure we have,
S = Starting point of car where angle of depression was $ 30^\circ $
R = Final point of the car where angle becomes $ 60^\circ $
 $ PQ = 100m $
And RS is the distance covered by car.
Let RS distance be x.
In triangle $ \vartriangle PQR $ , taking tan of angle of depression we get,
 $ \tan {\theta _2} = \dfrac{{PQ}}{{QR}} $ (As the tangent of angle in a right angle triangle is the ratio of its perpendicular and base.)
Putting the value of PQ and $ {\theta _2} $ in the above equation we get,
 $ \tan 60^\circ = \dfrac{{100}}{{QR}} $
As we know that the value of $ \tan 60^\circ = \sqrt 3 $ , putting this in the above equation,
 $ \sqrt 3 = \dfrac{{100}}{{QR}} $
By cross multiplication we get,
 $ QR = \dfrac{{100}}{{\sqrt 3 }} $
In triangle $ \vartriangle PQS $ , taking tan of angle of depression we get,
 $ \tan {\theta _1} = \dfrac{{PQ}}{{QS}} $ (As the tangent of angle in a right angle triangle is the ratio of its perpendicular and base.)
We can write QS as the sum of the QR and RS,
 $ \tan {\theta _1} = \dfrac{{PQ}}{{QR + RS}} $
Putting the value of PQ, RS, QR and $ {\theta _1} $ in the above equation we get,
 $ \tan 30^\circ = \dfrac{{100}}{{\dfrac{{100}}{{\sqrt 3 }} + x}} $
As we know that the value of $ \tan 30^\circ = \dfrac{1}{{\sqrt 3 }} $ , putting this in the above equation,
 $ \dfrac{1}{{\sqrt 3 }} = \dfrac{{100}}{{\dfrac{{100}}{{\sqrt 3 }} + x}} $
By cross multiplication we get,
 $ \dfrac{{100}}{{\sqrt 3 }} + x = 100\sqrt 3 $
Simplifying the above equation we get,
 $ x = 100\sqrt 3 - \dfrac{{100}}{{\sqrt 3 }} $
Taking L.C.M.,
 $ x = \dfrac{{(100 \times 3) - 100}}{{\sqrt 3 }} = \dfrac{{300 - 100}}{{\sqrt 3 }} = \dfrac{{200}}{{\sqrt 3 }} $
Hence, $ RS = \dfrac{{200}}{{\sqrt 3 }} $
Multiplying $ \sqrt 3 $ in both numerator and denominator we get,
 $ RS = \dfrac{{200\sqrt 3 }}{3} $
i.e. The distance (In meters) travelled by the car during this time is $ \dfrac{{200\sqrt 3 }}{3} $ meters.

Option B is correct.

Note: We can also do this in an alternative method in which first we will derive distance QS by taking tangent of depression angle at S. Then we can derive the distance QR accordingly at point R. By subtracting QR distance from QS, we will get the distance travelled by the car.
Angle of depression is the downwards angle from the horizontal to a line of sight from observer to some point of interest.
One angle can be the angle of depression when you are measuring the angle from the top of the perpendicular and it can be the angle of elevation when you are measuring from the ground.
Tan of the angle of the right angled triangle is the ratio of perpendicular (side that is in front of that angle) to the base.