A ladder is placed against a wall such that it just reaches the top of the wall. The foot of the ladder is at a distance of 5 metres from the wall. The angle of elevation of the top of the wall from the base of the ladder is 15°. What is the length (in metres) of the ladder?

Answer: Option B

Solution

Answer: Option B

Solution:

$$\begin{array}{rl}& \cos 15^{\circ }=\frac{BC}{AC}\\ & \frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{5}{AC}\\ & \frac{\sqrt{6}+\sqrt{2}}{2\times 2}=\frac{5}{AC}\\ & \therefore AC=\frac{20}{\sqrt{6}+\sqrt{2}}\\ & AC=\frac{20\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\\ & AC=\frac{20}{4}\left(\sqrt{6}-\sqrt{2}\right)\\ & AC=5\sqrt{6}-5\sqrt{2}\end{array}$$