Q16 of 35 Page 219

A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change its measure from 30 to 45, how soon after this, will the car reach the tower?

Let AB be the tower with height of tower AB = h



At C the angle of depression of car measures 30 and 12 minutes later it reaches D where angle of depression is 45.


Let CD = x, D = y


Here, AB = h, ∠ACB = ∠XAC = 30⁰


∠ADB = ∠XAD = 45⁰


In ∆ACB,




x + y = √3h ……(1)


In ∆ABD,




H = y ………(2)


From equation 1 and 2,


x + y = √3y


x = (√3 – 1)y


The distance covered by car in 12 minutes is CD = x


So, time taken to cover distance x is 12 minutes.


So, y = Time taken to cover distance DB







= 6 × 2.73


= 16.38 minutes


The time taken by car to reach the tower is 16.38 minutes.


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The angles of elevation of the top of a tower from two points at distance a and b metres from the base and in the same straight line with it are complementary. Prove that the height of the tower is metres.

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If the angle of elevation of a cloud from a point h metres above a lake has measure a and the angle of depression of its reflection in the lake has measure β, prove that the height of the h(tani3 + tana) cloud is .

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From the top of a building , 60 m high, the angles of depression of the top and bottom at a vertical lamp post are observed to have measure 30 and 60 respectively. Find,

(1) the horizontal distance between building and lamp post.


(2) the height of the lamp post.


(3) the difference between the heights of the building and the lamp post.