Question : 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 the ladder makes an angle of 30° with the ground. How much did the ladder slip from the top of the wall?
Option 1: 12 ft
Option 2: 20 ft
Option 3: 7.32 ft
Option 4: 18 ft
Correct Answer: 7.32 ft
Solution :
Given:
The foot of the ladder is 10 ft away from the wall, and the ladder is making an angle of 60° with the ground.
We know the trigonometric ratios, $\tan \theta = \frac{\text{Height}}{\text{Base}}$ and $\sin \theta = \frac{\text{Height}}{\text{Hypotenuse}}$
Here, $AC = DE =$ height of the ladder
In $\triangle ACB$, $\tan60°=\frac{AB}{BC}$
$\sqrt3=\frac{AB}{10}$
⇒ $AB=10\sqrt3$ ft
Using Pythagoras theorem in $\triangle ABC$, $(AC)^2=(AB)^2+(BC)^2$
⇒ $AC=\sqrt{(AB)^2+(BC)^2}$
⇒ $AC=\sqrt{(10\sqrt3)^2+(10)^2}$
⇒ $AC=\sqrt{300+100}$
⇒ $AC=\sqrt{400}$
⇒ $AC=20$ ft
⇒ $AC = DE =$ 20 ft
In $\triangle BDE$, $\sin 30°=\frac{BE}{DE}$
⇒ $\frac{1}{2}=\frac{BE}{20}$
⇒ $BE=10$ ft
So, $AE=(10\sqrt3-10)$
⇒ $10(1.732-1)=10\times 0.732 = 7.32$ ft
Hence, the correct answer is 7.32 ft.
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