Question

a straight highway leads to the foot of a tower. a man standing on the top of the tower observes a car at an angle of depression of 30° ,which is approaching the foot of the tower with a uniform speed. six seconds later, the angle of depression of the car is found to be 60° . find the time taken by the car to reach the foot of the tower from the point.

Answer 1

Answer:

Step-by-step explanation:

Lt AB is the tower and BD is the highway.

Nw frm Δ ADB,

     tan30 = AB/BD

= 1/√3 = AB/BD

= AB = BD/√3    .............1

Again frm Δ ACB

   tan60 = AB/BC

= √3 = AB/BC

= AB = BC√3 ........2

frm en= 1 and 2

BD/√3 = BC√3

= (BC + CD)/√3 = BC√3

= BC + CD = BC√3*√3

= BC + CD = 3BC

= 3BC - BC = CD

= 2BC = CD

= BC = CD/2  

time taken by car to cover CD = 6 Sec

So time taken by car to cover BC = 6/2 = 3 sec

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Answer 2

Question:

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

see above attachment