Q. 144.2( 75 Votes )

# A straight highway leads to the foot of a tower. A man standing on the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower form this point.

In the figure, let AB be the tower and D be the position of the car. Let C be the position of the car 6seconds later. The angles of depression are DAE and CAE. Join B, C, and C, D and A, E. Since AE is parallel to BD, we must have, ADB = DAE = 30°, CAE = ACB = 60°.
We get two right-angled triangles ∆ABC and ∆ABD, both right angled at B. We use trigonometric ratio tan for both the triangles using BC as base and AB as height for ∆ABC and BD as base and AB as height in ∆ABD. We find the CD.

Let BC = x.

In ∆ABC,

or,

Again, in ∆ABD,

or,

CD = 2x

Now, the speed of the car = CD/6 = 2x/6 = x/3.

Hence, time taken by the car to reach the foot of the tower is

= BC/(Speed of the car) = 3 secs.

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