The bearing of Q from P is 050^ and the bearing of R from Q is 130^. Given |PQ|=8 km and |QR|=13 km, find the length of PR to 3 significant figures.

The bearing of Q from P is $050^\circ$. This means the angle from the North line at P, clockwise to PQ, is $50^\circ$.
Since the North lines at P and Q are parallel, the alternate interior angle to the $50^\circ$ bearing at P is the angle between the North line at Q and QP. This angle is also $50^\circ$.
The bearing of R from Q is $130^\circ$. This is the angle from the North line at Q, clockwise to QR.
The interior angle $\angle PQR$ is the angle between QP and QR.

$$\angle PQR = 130^\circ - 50^\circ = 80^\circ$$

We have a triangle PQR with two sides and the included angle:
$|PQ| = 8 \text{ km}$
$|QR| = 13 \text{ km}$
$\angle PQR = 80^\circ$
Using the Cosine Rule: $c^2 = a^2 + b^2 - 2ab \cos C$
Here, $PR^2 = PQ^2 + QR^2 - 2(PQ)(QR) \cos(\angle PQR)$

$$PR^2 = 8^2 + 13^2 - 2(8)(13) \cos(80^\circ)$$

$$PR^2 = 64 + 169 - 208 \cos(80^\circ)$$ $$PR^2 = 233 - 208(0.173648...)$$ $$PR^2 = 233 - 36.1208...$$ $$PR^2 = 196.8791...$$

$$PR = \sqrt{196.8791...}$$ $$PR = 14.0313... \text{ km}$$

Rounding to 3 significant figures:

$$PR \approx 14.0 \text{ km}$$

The length of PR is $\text{14.0 km}$.