Question

There are two windows on the wall of a building that need repairs. A ladder 30 m long is placed against a wall such that it just reaches the first window which is 26 m high. The foot of the ladder is at point A. After the first window is fixed, the foot of the ladder is pushed backwards to point B so that the ladder can reach the second window. The angle made by the ladder with the ground is reduced by half, as a result of pushing the ladder. The distance between points A and B is:

• A. $< 9$ m
• B. $\geq 9$ m and $< 9.5$ m
• C. $\geq 9.5$ m and $< 10$ m
• D. $\geq 10$ m and $< 10.5$ m
• E. $\geq 10.5$ m

Hint:

In ladder problems, use trigonometric identities like \(\sin\theta, \cos\theta\) and Pythagoras. For angle halving, always recalculate the new base carefully.

Answer 1

Answer: (E)

Step 1: First position of ladder (point (A)
The ladder length is 30 m, vertical height = 26 m.
By Pythagoras: \[ \text{Base} = \sqrt{30^2 - 26^2} = \sqrt{900 - 676} = \sqrt{224} = 14.97 \, \text{m Approx.)} \] So, distance OA = 14.97 m.
Angle with ground: \[ \theta = \arcsin\left(\frac{26}{30}\right) = \arcsin\left(\frac{13}{15}\right) \] Step 2: Second position (point (B)
Angle is halved \(⇒\) new angle = \(\tfrac{\theta}{2}\).
Using formula for ladder base: \[ \text{New base} = 30 \cos\left(\frac{\theta}{2}\right) \] Since \(\cos(\thet(A) = \tfrac{14.97}{30} = 0.499\), \[ \theta \approx 60^\circ \] \[ \frac{\theta}{2} \approx 30^\circ \] \[ \text{New base} = 30 \cos(30^\cir(C) = 30 \times \frac{\sqrt{3}}{2} = 25.98 \, \text{m} \] Step 3: Distance AB \[ AB = \text{New base} - \text{Old base} = 25.98 - 14.97 = 11.01 \, \text{m} \] \[ \boxed{AB \geq 10.5 \, \text{m}} \]