In this triangle: - AB = 9 cm - AC = 10 cm - BC > 5 cm - angle BCA = 60° - angle ABC < 90° - OCR - GCSE Maths - Question 19 - 2018 - Paper 6

Using the Sine Rule:

$\frac{AB}{\sin(ABC)} = \frac{AC}{\sin(BCA)}$

Rearranging for sin(ABC):

$\sin(ABC) = \frac{AB \times \sin(BCA)}{AC}$

Substituting the values:

$\sin(ABC) = \frac{9 \times \sin(60°)}{10} = \frac{9 \times \frac{\sqrt{3}}{2}}{10} = \frac{9\sqrt{3}}{20}$

The area of triangle ABC can be calculated using:

$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(BCA)$

Substituting the values:

$\text{Area} = \frac{1}{2} \times 9 \times 10 \times \sin(60°) = \frac{1}{2} \times 9 \times 10 \times \frac{\sqrt{3}}{2} = \frac{90\sqrt{3}}{4} = 22.5\sqrt{3} \approx 38.97 ext{ cm}^2$