Question

From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45°, respectively. Find the distance between the cars. [Take √3 = 1.73]

Answer 1

In the figure,

A and C are the positions of the cars.

BD = 100 m.

The angles of depression are ∠DAB and ∠DCB.

Since the line XY is parallel to AC, we have

∠YDC = ∠DCB = 30° and

∠XDA = ∠DAB = 45°

To find: Distance between the two cars

From right ∆ABD:

tan45° = BD/AB

1 = 100/AB

or AB = 100

From right ∆CBD:

tan 30° = BD/BC

1/√3 = 100/BC

or BC = 100√3

Now, AC = AB + BC

= 100 + 100√3

= 100(1 + √3)

= 273.2

Hence distance between the cars is 273.2 meters.