In die diagram hieronder is T 'n haak in die plafon van 'n kunsgalery - NSC Mathematics - Question 8 - 2021 - Paper 2

Applying the sine rule in triangle QST:

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

Where TS = QS = 5 tan x, substituting gives:

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

Using the identity $sin(180° - θ) = sin(θ)$:

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

We know that $sin(2x) = 2 sin(x) cos(x)$, thus:

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

This simplifies to:

$QT = \frac{5 \cdot 1}{2 cos(x)} = 10 sin x$

To find the area of triangle ΔTQR, we use the formula:

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

Where a and b are lengths QT and QR, and C is angle TQR:

Substituting QT = 10 sin(25°), QR = 5, and angle TQR = 70°:

$Area = \frac{1}{2} \cdot (10 sin(25°)) \cdot 5 \cdot sin(70°)$

Calculating further:

$Area = \frac{50}{2} \cdot sin(25°) \cdot sin(70°)$

Thus, the area:

$Area ≈ 9.93 unit^2$