1. **State the problem:** We are given two angles of a triangle, 40° and 49°, and one side length, 13. We need to find the other sides of the triangle. 2. **Recall the triangle angle sum rule:** The sum of angles in a triangle is 180°. So, the third angle is $$180^\circ - 40^\circ - 49^\circ = 91^\circ$$. 3. **Use the Law of Sines:** For a triangle with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively, $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$. 4. **Assign known values:** Let side $c = 13$ be opposite angle $C = 91^\circ$. We want to find sides $a$ and $b$ opposite angles $A = 40^\circ$ and $B = 49^\circ$. 5. **Calculate side $a$:** $$a = c \times \frac{\sin A}{\sin C} = 13 \times \frac{\sin 40^\circ}{\sin 91^\circ}$$ Since $\sin 91^\circ \approx 1$, $$a \approx 13 \times \sin 40^\circ = 13 \times 0.6428 = 8.36$$ 6. **Calculate side $b$:** $$b = c \times \frac{\sin B}{\sin C} = 13 \times \frac{\sin 49^\circ}{\sin 91^\circ}$$ Again, $\sin 91^\circ \approx 1$, $$b \approx 13 \times \sin 49^\circ = 13 \times 0.7547 = 9.81$$ 7. **Final answer:** The sides opposite the 40° and 49° angles are approximately 8.36 and 9.81 respectively.