iGCSE Mathematics (0580) - C6.2 Right-angled triangles- Exam Style Questions Paper 3- New Syllabus
▶️ Answer/Explanation
Using the SOH CAH TOA rules, side BC is the hypotenuse, and side AB ($32$ cm) is the side opposite to angle C ($27^\circ$).
$\sin(27^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{32}{BC}$
$BC = \frac{32}{\sin(27^\circ)} = 70.485…$
✅ Answer: $70.5$ cm
▶️ Answer/Explanation
First, use Pythagoras’ theorem to find the shared side $BD$ in the first triangle: $BD = \sqrt{14.7^2 – 10.8^2} \approx 9.97$ m.
Now, look at the second triangle (BCD). You know the adjacent side ($BD = 9.97$ m) and want to find the opposite side ($d$) to the $52^{\circ}$ angle.
Using the tangent ratio: $\tan(52^{\circ}) = \frac{d}{BD} \Rightarrow d = 9.97 \times \tan(52^{\circ}) \approx 12.76…$
✅ Answer: $12.8$ m