1. Problem statement: P is 12 km due north of Q, the bearing of R from P is 135^\circ and from Q is 120^\circ; find (i) angle PRQ and (ii) distance PR. 2. Set a coordinate frame: place Q at (0,0) and P at (0,12) so north is the positive y-direction. 3. Formula for converting a bearing $\beta$ (measured clockwise from north) to a unit direction vector: $$\mathbf{u}(\beta)=(\sin\beta,\;\cos\beta).$$ 4. Apply the formula to the two bearings to get unit direction vectors from P and Q toward R. $\mathbf{u}_P=(\sin135^\circ,\;\cos135^\circ)=(\frac{\sqrt{2}}{2},\; -\frac{\sqrt{2}}{2}).$ $\mathbf{u}_Q=(\sin120^\circ,\;\cos120^\circ)=(\frac{\sqrt{3}}{2},\; -\frac{1}{2}).$ 5. Parametric equations of the two lines PR and QR with parameters $t,s\ge 0$ (distance along each direction): $$R=P+t\mathbf{u}_P=(0,12)+t\Big(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\Big).$$ $$R=Q+s\mathbf{u}_Q=(0,0)+s\Big(\frac{\sqrt{3}}{2},-\frac{1}{2}\Big).$$ 6. Equate coordinates to find $t,s$. From the x-coordinates: $$t\frac{\sqrt{2}}{2}=s\frac{\sqrt{3}}{2}\Rightarrow t\sqrt{2}=s\sqrt{3}\Rightarrow t=s\sqrt{\tfrac{3}{2}}.$$ From the y-coordinates: $$12-t\frac{\sqrt{2}}{2}=-s\frac{1}{2}.$$ Substitute $t=s\sqrt{\tfrac{3}{2}}$ into the y-equation and solve for $s$. 7. Solve for $s$ (showing algebraic simplification with cancellation): $$s=\frac{24}{\sqrt{3}-1}=\frac{24(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{24(\sqrt{3}+1)}{2}=\frac{\cancel{24}(\sqrt{3}+1)}{\cancel{2}}=12(\sqrt{3}+1).$$ 8. Find $t$ (which equals PR because $t$ multiplies a unit vector): From $t\sqrt{2}=s\sqrt{3}$ we get $$t=\frac{s\sqrt{3}}{\sqrt{2}}=12(\sqrt{3}+1)\frac{\sqrt{3}}{\sqrt{2}}=\frac{12(3+\sqrt{3})}{\sqrt{2}}=\frac{12(3+\sqrt{3})\sqrt{2}}{2}=6(3+\sqrt{3})\sqrt{2}=18\sqrt{2}+6\sqrt{6}.$$ So the exact distance is $$PR=18\sqrt{2}+6\sqrt{6}\text{ km}.$$ Numerical approximation: $PR\approx 40.15\text{ km}$. 9. Find angle PRQ (the angle at R between RP and RQ). The angle between the two direction unit vectors $\mathbf{u}_P$ and $\mathbf{u}_Q$ is the same as angle PRQ (negation of both vectors does not change the angle). Use the dot product formula: $$\cos\angle PRQ=\mathbf{u}_P\cdot\mathbf{u}_Q=\sin135^\circ\sin120^\circ+\cos135^\circ\cos120^\circ=\frac{\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{6}+\sqrt{2}}{4}.$$ Thus $$\angle PRQ=\cos^{-1}\!\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)=15^\circ.$$ 10. Answers: (i) $\angle PRQ=15^\circ$. (ii) $PR=18\sqrt{2}+6\sqrt{6}\approx 40.15\text{ km}$.