Question

Angle of depression from tower top to A is 30°. After 20 seconds at B, it is 45°. Time taken from B to reach the base is :

• A. 10
• B. 10(√3 + 1)
• C. 10√3
• D. 10(√3 - 1)

Hint:

In problems of uniform speed, time taken is proportional to distance: $T_1/D_1 = T_2/D_2$.

Answer 1

The Correct Option is B

To solve this problem, we need to use the concept of angles of depression and trigonometry. Angles of depression are measured from the horizontal downward to the line of sight. Let's break down the problem and find the necessary solution:

1. Consider a tower with height \( h \). From the top of the tower, the angle of depression to point A is \( 30^\circ \) and to point B is \( 45^\circ \).
2. Let's assume the distance from the base of the tower to point A is \( x \) and the distance to point B is \( y \).
3. Using the angle of depression to point A, we apply the tangent function:
   • \(\tan 30^\circ = \frac{h}{x}\)
   • \(\frac{1}{\sqrt{3}} = \frac{h}{x}\)
   • Therefore, \( x = h \cdot \sqrt{3} \).
4. Using the angle of depression to point B:
   • \(\tan 45^\circ = \frac{h}{y}\)
   • \(1 = \frac{h}{y}\)
   • Therefore, \( y = h \).
5. The time taken to move from A to B is 20 seconds. This corresponds to moving a horizontal distance of \( x - y \) (i.e., \( h\sqrt{3} - h \)).
   • \(x - y = h(\sqrt{3} - 1)\).
6. The speed \( v \) is then the distance divided by time:
   • \(v = \frac{h(\sqrt{3} - 1)}{20}\).
7. Now, we need to determine the time taken to travel from B to the base of the tower, covering the distance \( y = h \). Time is given by distance divided by speed:
   • \(t = \frac{y}{v} = \frac{h}{\frac{h(\sqrt{3} - 1)}{20}} = \frac{20}{\sqrt{3} - 1}\).
8. To rationalize the denominator, multiply numerator and denominator by \(\sqrt{3} + 1\):
   • \(t = \frac{20(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{20(\sqrt{3} + 1)}{2} = 10(\sqrt{3} + 1)\).
9. Thus, the time taken from point B to the base of the tower is \(10(\sqrt{3} + 1)\).

The correct answer is 10(√3 + 1).