Two ladders, $PS$ and $QT$, are placed against a vertical wall $TR$. $PS = 10$ m and $QT = 12$ m. The ladder $PS$ makes an angle of $60^\circ$ with the horizontal. The distance $PQ = 0.5$ m. We need to find: i) The angle which $QT$ makes with the horizontal. ii) The length of point $T$ above the horizontal. The answer must be correct to two significant figures.

1. Problem Description

Two ladders, $PS$ and $QT$, are placed against a vertical wall $TR$. $PS = 10$ m and $QT = 12$ m. The ladder $PS$ makes an angle of $60^\circ$ with the horizontal. The distance $PQ = 0.5$ m. We need to find:

i) The angle which $QT$ makes with the horizontal.

ii) The length of point $T$ above the horizontal. The answer must be correct to two significant figures.

2. Solution Steps

i) Let $\theta$ be the angle that $QT$ makes with the horizontal.

We know that $PQ = 0.5$ m. Also, $PT = PQ + QT \cos \theta$.

Since $\angle SPR = 60^\circ$, we have $\cos 60^\circ = \frac{PR}{PS}$.

Therefore, $PR = PS \cos 60^\circ = 10 \times \frac{1}{2} = 5$ m.

Also, $QR = PR - PQ = 5 - 0.5 = 4.5$ m.

Now, we have $\cos \theta = \frac{QR}{QT} = \frac{4.5}{12} = \frac{3}{8} = 0.375$.

Therefore, $\theta = \arccos(0.375) \approx 67.9759^\circ$.

Correct to two significant figures, $\theta = 68^\circ$.

ii) We need to find the length of $TR$.

We know that $\sin 60^\circ = \frac{SR}{PS}$.

Therefore, $SR = PS \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$ m.

Also, $\sin \theta = \frac{TR}{QT}$.

Therefore, $TR = QT \sin \theta = 12 \sin(67.9759^\circ) \approx 12 \times 0.9272 = 11.1264$.

$TR = 12 \sin(67.9759^\circ)$

$TR \approx 11.13$ m.

Correct to two significant figures, $TR = 11$ m.

3. Final Answer

i) The angle which $QT$ makes with the horizontal is $68^\circ$.

ii) The length of point $T$ above the horizontal is $11$ m.