A ladder is placed along a wall such that its upper end is touching the top of the wall. The foot of the ladder is 10 ft away from the wall and the ladder is making an angle of 60° with the ground. When a man starts climbing on it, it slips and now ladder makes an angle of 30° with ground. How much did the ladder slip?

Answer: Option A

Solution

Answer: Option A

Solution:

$$\begin{array}{rl}& AB=10\tan 60^{\circ }=10\sqrt{3}\\ & BC=10\sec 60^{\circ }=20\end{array}$$
Since, length of ladder will remain same, hence BC = DE = 20
$$\begin{array}{rl}& \text{In }\Delta ADE,DE=20\text{ ft}\\ & \Rightarrow \frac{AD}{DE}=\cos 30^{\circ }\\ & \Rightarrow AD=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}\\ & \Rightarrow x=AD-AC\\ & \Rightarrow x=10\sqrt{3}-10\\ & \Rightarrow x=10\left(\sqrt{3}-1\right)\text{ft}\end{array}$$