Q. 134.2( 6 Votes )

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 :


In the figure, let AC be the tower of height 100m. The angles of depression are BAE and DAF. Given, AC = 100m. Since the line EF is parallel to the line BD, we have BAE = BAC = 30°, DAF = DAC = 45°, where B, D are the positions of the cars. Join B, C, and C, D. We get two right-angled triangles ∆ABC and ∆ADC, both right angled at C. We use trigonometric ratio tan for both the triangles, where BC is the base and Ac is the height (for ∆ABC) and CD is the base and AC is the height in ∆ACD. We are to find the distance between the two cars, i.e. BD.

In ∆ABC,


In ∆ADC,


DC = 100

Therefore, the distance between the two cars is

BD = BC + CD = 100√3 + 100 = 100(√3 + 1) = 273 m.

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