Question 56
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?
Solution
So, $$\tan15^{\circ\ }=\frac{AB}{BC}=\frac{AB}{5}\ .$$
We know that , $$\tan15^{\circ\ }=\left(2-\sqrt{3}\right).$$
So, $$AB=\ 5\left(2-\sqrt{3}\right)\ .$$
So, Ladder height = AC = $$\sqrt{BC^2+AB^2}.$$
Now, AC = $$\sqrt{5^2+\left(5\left(2-\sqrt{3}\right)\right)^2}=\sqrt{25+100-100\sqrt{3}+75}=\sqrt{\left(5\sqrt{6}-5\sqrt{2}\right)^2}=\left(5\sqrt{6}-5\sqrt{2}\right).$$
B is correct choice.
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