In ABC, A=40^, BC=12 and AB=15. Determine the other angles of the triangle. Assume that C is acute. In PQR, Q=110^, R=40^ and PQ=13. Calculate the lengths of the sides.

The Sine Rule states that for a triangle with sides $a, b, c$ and opposite angles $\hat{A}, \hat{B}, \hat{C}$:

$$\frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}} = \frac{c}{\sin \hat{C}}$$

Given $\hat{A} = 40^\circ$, $BC = 12$ (side $a$), and $AB = 15$ (side $c$). We want to find $\hat{C}$.

$$\frac{BC}{\sin \hat{A}} = \frac{AB}{\sin \hat{C}}$$ $$\frac{12}{\sin 40^\circ} = \frac{15}{\sin \hat{C}}$$ $$\sin \hat{C} = \frac{15 \sin 40^\circ}{12}$$ $$\sin \hat{C} = \frac{15 \times 0.6427876}{12}$$ $$\sin \hat{C} \approx 0.8034845$$

Since $\hat{C}$ is acute, we take the principal value:

$$\hat{C} = \arcsin(0.8034845)$$ $$\hat{C} \approx 53.46^\circ$$

The sum of angles in a triangle is $180^\circ$.

$$\hat{A} + \hat{B} + \hat{C} = 180^\circ$$ $$40^\circ + \hat{B} + 53.46^\circ = 180^\circ$$ $$\hat{B} = 180^\circ - 40^\circ - 53.46^\circ$$ $$\hat{B} = 180^\circ - 93.46^\circ$$ $$\hat{B} \approx 86.54^\circ$$

The other angles of the triangle are $\boxed{\hat{C} \approx 53.46^\circ, \hat{B} \approx 86.54^\circ}$.

Given $\hat{Q} = 110^\circ$ and $\hat{R} = 40^\circ$.

$$\hat{P} + \hat{Q} + \hat{R} = 180^\circ$$ $$\hat{P} + 110^\circ + 40^\circ = 180^\circ$$ $$\hat{P} = 180^\circ - 110^\circ - 40^\circ$$ $$\hat{P} = 180^\circ - 150^\circ$$ $$\hat{P} = 30^\circ$$

Given $PQ = 13$. We want to find $QR$ (opposite $\hat{P}$) and $PR$ (opposite $\hat{Q}$).

$$\frac{PQ}{\sin \hat{R}} = \frac{QR}{\sin \hat{P}}$$ $$\frac{13}{\sin 40^\circ} = \frac{QR}{\sin 30^\circ}$$ $$QR = \frac{13 \sin 30^\circ}{\sin 40^\circ}$$ $$QR = \frac{13 \times 0.5}{0.6427876}$$ $$QR = \frac{6.5}{0.6427876}$$ $$QR \approx 10.12$$

$$\frac{PQ}{\sin \hat{R}} = \frac{PR}{\sin \hat{Q}}$$ $$\frac{13}{\sin 40^\circ} = \frac{PR}{\sin 110^\circ}$$ $$PR = \frac{13 \sin 110^\circ}{\sin 40^\circ}$$ $$PR = \frac{13 \times 0.9396926}{0.6427876}$$ $$PR = \frac{12.2160038}{0.6427876}$$ $$PR \approx 19.00$$

The lengths of the sides are $\boxed{QR \approx 10.12, PR \approx 19.00}$.