1. **Problem statement:** In triangle $\triangle PQR$, given $\angle Q = 42^\circ$, side $PR = 12$ cm, and side $PQ = 10.2$ cm, find: (i) $\angle R$ (ii) $\angle P$ (iii) length of side $QR$ 2. **Step 1: Find $\angle R$ using the sine rule.** The sine rule states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ Here, label sides opposite to angles: $a = QR$, $b = PR = 12$, $c = PQ = 10.2$. Given $\angle Q = 42^\circ$, side opposite $Q$ is $PR = 12$ cm. We want to find $\angle R$ opposite side $PQ = 10.2$ cm. Using sine rule between angles $Q$ and $R$: $$\frac{PR}{\sin Q} = \frac{PQ}{\sin R} \implies \frac{12}{\sin 42^\circ} = \frac{10.2}{\sin R}$$ Rearranged: $$\sin R = \frac{10.2 \times \sin 42^\circ}{12}$$ Calculate: $$\sin 42^\circ \approx 0.6691$$ $$\sin R = \frac{10.2 \times 0.6691}{12} = \frac{6.823}{12} = 0.5686$$ Find $\angle R$: $$R = \sin^{-1}(0.5686) \approx 34.7^\circ$$ 3. **Step 2: Find $\angle P$.** Sum of angles in a triangle is $180^\circ$: $$P = 180^\circ - Q - R = 180^\circ - 42^\circ - 34.7^\circ = 103.3^\circ$$ 4. **Step 3: Find length $QR$ opposite $P$.** Using sine rule again between $P$ and $Q$: $$\frac{QR}{\sin P} = \frac{PR}{\sin Q}$$ Rearranged: $$QR = \frac{PR \times \sin P}{\sin Q} = \frac{12 \times \sin 103.3^\circ}{\sin 42^\circ}$$ Calculate sines: $$\sin 103.3^\circ \approx 0.9744$$ $$\sin 42^\circ \approx 0.6691$$ Calculate $QR$: $$QR = \frac{12 \times 0.9744}{0.6691} = \frac{11.693}{0.6691} \approx 17.47 \text{ cm}$$ **Final answers:** (i) $\angle R \approx 34.7^\circ$ (ii) $\angle P \approx 103.3^\circ$ (iii) $QR \approx 17.5$ cm (to 3 significant figures)