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

Using the sine rule again in triangle QTS:

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

Substituting TS = 5 tan x:

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

Using the identity for sin(2x) = 2sin(x)cos(x), we have:

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

The area of triangle ΔTQR can be calculated using:

$Area = \frac{1}{2} \cdot QT \cdot QR \cdot sin(TQR)$

Substituting the values we found:

• We have already determined that QT = 10 sin(x) when x = 25°: $QT = 10 \cdot sin(25°)$
• And QR = 5.

Thus, the area becomes:

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

Calculating this will give the final area.