**Q) **From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angle of depression 30^{0} and 45^{0} respectively. Find the distance between the two cars. (use 3 = 1.73)

**Ans: **

Let PQ be the tower, A and B the two cars, man observes the 2 cars from point P. Let angle of elevations and distances of cars from the tower be D_{1} and D_{2} as shown in the image above.

In Δ APQ, tan 30 =

** D _{1} = 100 **

**√3 = 100 x 1.73 = 173 m**

In Δ BPQ, tan 45 =

** D _{2} = 100 **

**m**

Hence, the distance between the both cars = D_{1} + D_{2} = 173 + 100 = 273 m

**Therefore, the distance between the two cars is 27****3 m.**