In the diagram below, T is a hook on the ceiling of an art gallery - NSC Mathematics - Question 8 - 2021 - Paper 2

In triangle $SQT$, we can apply the sine rule again:

Using the angle $180° - 2x$ at vertex $S$:

$\frac{QT}{TS} = \frac{\sin(180° - 2x)}{\sin x}$

Since $TS = QS = 5 \tan x$, substituting gives us:

$QT = 5 \tan x \cdot \frac{\sin(180° - 2x)}{\sin x}$

Knowing $\sin(180° - 2x) = \sin 2x$:

$QT = 5 \tan x \cdot \frac{\sin 2x}{\sin x}$

From the double angle identity: $\sin 2x = 2 \sin x \cos x$, we substitute:

$QT = 5 \tan x \cdot \frac{2 \sin x \cos x}{\sin x}$

So,

$QT = 10 \tan x \cdot \cos x$

Since $\tan x = \frac{\sin x}{\cos x}$, this simplifies to:

$QT = 10 \sin x$

To find the area of triangle $QRT$, we can use the formula:

$\text{Area} = \frac{1}{2} a b \sin(C)$

Where $a$ and $b$ are the lengths of sides and $C$ is the angle between them. Here,

• $a = QT = 10 \sin x$
• $b = QR = 5$
• $C = \angle TQR + \angle RQT = 70° + 25° = 95°$.

Thus, the area is:

$\text{Area} = \frac{1}{2} \cdot (10 \sin x) \cdot (5) \cdot \sin(95°)$

Calculating gives:

$\text{Area} = 25 \sin x \sin(95°)$

Approximating $\sin(95°)$ as approximately 1, when $x = 25°$ makes:

$\text{Area} = 25 \cdot \sin(25°) \cdot 1 = 25 \cdot 0.4226 = 10.565$

So, thus yielding:

$ext{Area} \approx 9.93 \text{ units}^2$