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 300 and 450 respectively. Find the distance between the two cars. (use 3 = 1.73)


From the top of a tower CBSE 10th Board important question

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 D1 and D2 as shown in the image above.

In Δ APQ, tan 30 = \frac{PQ}{AQ}

\frac{1}{\sqrt 3} = \frac{100}{D_1}

\therefore D1 = 100 √3  = 100 x 1.73 = 173 m

In Δ BPQ, tan 45 = \frac{PQ}{BQ}

1 = \frac{100}{D_2}

\therefore D2 = 100 m

Hence, the distance between the both cars = D1 + D2 = 173 + 100 = 273 m

Therefore, the distance between the two cars is 273 m.

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