Solve the equation x(x + 3) = (x − 1)(2x − 1) by general formula.

Jumla ya pembe katika pembetatu ni $180^\circ$.

$$\angle PRQ = 180^\circ - (\angle RPQ + \angle PQR)$$ $$\angle PRQ = 180^\circ - (84^\circ + 67^\circ)$$ $$\angle PRQ = 180^\circ - 151^\circ$$ $$\angle PRQ = 29^\circ$$

Kanuni ya Saini inasema $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
Tunajua PQ = 3 km, $\angle RPQ = 84^\circ$, $\angle PQR = 67^\circ$, na $\angle PRQ = 29^\circ$.
Ili kupata PR (upande unaopingana na $\angle PQR$), tunatumia:

$$\frac{PR}{\sin(\angle PQR)} = \frac{PQ}{\sin(\angle PRQ)}$$ $$\frac{PR}{\sin(67^\circ)} = \frac{3}{\sin(29^\circ)}$$ $$PR = \frac{3 \times \sin(67^\circ)}{\sin(29^\circ)}$$ $$PR = \frac{3 \times 0.9205}{0.4848}$$ $$PR = \frac{2.7615}{0.4848}$$ $$PR \approx 5.696 \text{ km}$$

Ili kupata QR (upande unaopingana na $\angle RPQ$), tunatumia:

$$\frac{QR}{\sin(\angle RPQ)} = \frac{PQ}{\sin(\angle PRQ)}$$ $$\frac{QR}{\sin(84^\circ)} = \frac{3}{\sin(29^\circ)}$$ $$QR = \frac{3 \times \sin(84^\circ)}{\sin(29^\circ)}$$ $$QR = \frac{3 \times 0.9945}{0.4848}$$ $$QR = \frac{2.9835}{0.4848}$$ $$QR \approx 6.154 \text{ km}$$

Majibu ya mwisho ni:

$$\boxed{PR \approx 5.70 \text{ km}, QR \approx 6.15 \text{ km}}$$