Question

A straight highway leads to the foot of a tower. A man standing at the top of the tower observe a car at an angle of depression of 30° , which is approaching the foot of the tower with uniform speed . Six Second 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 ......


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Answer 1

Here is ur soln ma'am...
hope it helps

Answer 2

From the figure,

Let AB be the height of tower, and C be the initial position of the car when angle of depression is 30°.

After 6 seconds, the car reaches to B such that angle of depression is 60°.

Now,

(i) In ΔADB,

⇒ (AB/DB) = tan 60

⇒ (AB/DB) = √3

⇒ AB = √3 * DB

⇒ DB = AB/√3.    -------- (1)

(ii) In ΔABC,

⇒ AB/BC = tan 30

⇒ (AB/BD + DC) = 1/√3

⇒ AB√3 = BD + DC

⇒ AB√3 = (AB/√3) + DC

⇒ 3AB = AB + √3 * DC

⇒ 3AB - AB = √3 * DC

⇒ 2AB = √3 * DC

⇒ DC = 2AB/√3.

So,

Time taken by car to travel Distance DB :

$= > \frac{6}{\frac{2AB}{\sqrt{3}}} * \frac{AB}{\sqrt{3}}$

$= > \frac{6\sqrt{3} * AB}{2AB * \sqrt{3} }$

$= > \frac{6}{2}$

$= > 3$

Therefore, time taken by car to reach foot of the tower is 3 seconds.

Hope this helps!