Car Approaching a Tower: 30° to 60° in 6 Seconds | Class 10

Quick answer: Let the tower have height $h$ and the car move at speed $v$. When the angle of depression is 60°, the car is at distance $vt$ from the foot, so $h = \sqrt{3}\,vt$. Six seconds earlier, at 30°, it was at $v(t + 6)$, so $h = \frac{v(t + 6)}{\sqrt{3}}$. Equating gives $3t = t + 6$, so t = 3 seconds. The speed and height cancel out.

The problem

A straight highway leads to the foot of a tower. A man at the top of the tower sees a car approaching at an angle of depression of 30°. Six seconds later the angle of depression is 60°. Find the time the car takes to reach the foot of the tower from this point. This is a well-known NCERT question from Some Applications of Trigonometry.

 A (top of tower)
 |\ \
 | \  \
 h  \   \
 |60°\ 30°\
 B----D-----C
 foot  <-6 s->

• AB = height of the tower = $h$
• C = first position of the car (angle of depression 30°, so ∠ACB = 30°)
• D = position 6 seconds later (angle of depression 60°, so ∠ADB = 60°)
• The angles of depression at A equal the angles of elevation at C and D (alternate angles).

Step-by-step solution

Step 1: Write distances using speed. Let the speed be $v$ and the time from D to B be $t$ seconds.

$BD = vt, \qquad CD = 6v, \qquad BC = v(t + 6)$

Step 2: In right triangle ABD.

$\tan 60^\circ = \frac{AB}{BD} \quad \Rightarrow \quad \sqrt{3} = \frac{h}{vt}$

$h = \sqrt{3}\,vt \qquad \text{(1)}$

Step 3: In right triangle ABC.

$\tan 30^\circ = \frac{AB}{BC} \quad \Rightarrow \quad \frac{1}{\sqrt{3}} = \frac{h}{v(t + 6)}$

$h = \frac{v(t + 6)}{\sqrt{3}} \qquad \text{(2)}$

Step 4: Equate (1) and (2) and cancel $v$.

$\sqrt{3}\,t = \frac{t + 6}{\sqrt{3}}$

$3t = t + 6$

$t = 3$

Answer: The car takes 3 seconds to reach the foot of the tower from the second position.

A quick way to see it

From (1) and (2), $BC = 3 \times BD$. So $CD = BC - BD = 2 \times BD$, which means the remaining distance BD is half of CD. At constant speed, the remaining time is half of 6 seconds, which is 3 seconds.

This works whenever the angles are 30° and 60°, whatever the height of the tower or the speed of the car.

Summary table

Exam tips

• Draw the horizontal line from the top of the tower and mark the angles of depression, then mark the equal angles of elevation at C and D.
• Write the distances as speed × time; the speed will cancel.
• Use tan in both triangles, because the height and the horizontal distance are involved.
• State the final answer in words with the unit (seconds).

Common mistakes

• Putting the angles of depression at the wrong points. The 30° angle belongs to the farther position C.
• Using sin or cos instead of tan.
• Trying to find a numerical speed or height; they are not needed and cannot be found from the data.
• Answering 6 seconds or 9 seconds instead of the time from the second position to the tower.

Frequently Asked Questions

How long does the car take to reach the tower?

It takes 3 seconds from the point where the angle of depression is 60°.

Why does the speed of the car not matter?

Both distances are written as speed × time, so the speed appears on both sides of the equation and cancels out.

Why is the angle of elevation at the car equal to the angle of depression?

The horizontal line at the top of the tower is parallel to the road, so the angle of depression and the angle of elevation are alternate angles and are equal.

What is the shortcut for 30° and 60° problems?

With angles of 30° and 60°, the farther distance is three times the nearer one, so the distance still to go is half the distance already covered.

Which trigonometric ratio is used in this problem?

The tangent ratio, because it links the height of the tower to the horizontal distance of the car.