Question

A man observes a car from the top of a tower, which is moving towards
the tower with a uniform speed. If the angle of depression of the car
changes from 30° to 45° in 12 minutes, find the time taken by the car
now to reach the tower.
​

Answer 1

${\huge{\underline{\underline{\rm{QuestiOn:}}}}}$

A man observes a car from the top of a tower, which is moving towards the tower with a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.

${\huge{\underline{\underline{\rm{SolutiOn:}}}}}$

${\huge{\red{Given\:that}}}$

• ∠ADB=30° and ∠ACB= 45°

⟹Let us suppose the height of tower (AB) be h

speed of car "v" and time is " t".

In ∆ABC

tan45°=

${\huge\frac{AB}{BC}}$

$1 = \frac{h}{vt} \\ h = vt$

In ∆ABD

tan30° =

$⟹\frac{h}{12v + vt} \\ ⟹\frac{1}{3 {}^{ \frac{1}{2} } } = \frac{h}{12v + vt} \\ ⟹12v + vt = \sqrt{3} h \\⟹ 12v + vt = \sqrt{3} vt$

$⟹12v = ( \sqrt{3} - 1)vt \\ ⟹ \frac{12v}{vt} = \sqrt{3} - 1 \\ ⟹\frac{12}{t} = 0.73 \\⟹ t = \frac{12}{0.73}$

t= 16.43 minutes.

${\huge{\underline{\blue{Conclusion:}}}}$

Therefore, time taken to reach the tower is 16.43 minutes.