1. **State the problem:** We have a triangle with angles 96° and 25°, and the side opposite the 25° angle is 13 units. We need to find the side length opposite the 96° angle using the law of sines. 2. **Find the third angle:** The sum of angles in a triangle is 180°. $$ 180^\circ - 96^\circ - 25^\circ = 59^\circ $$ 3. **Set up the law of sines:** $$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} $$ Here, let $a$ be the side opposite 96°, $b=13$ opposite 25°, and $A=96^\circ$, $B=25^\circ$. 4. **Apply the law of sines to find $a$:** $$ \frac{a}{\sin 96^\circ} = \frac{13}{\sin 25^\circ} $$ 5. **Solve for $a$:** $$ a = \frac{13 \times \sin 96^\circ}{\sin 25^\circ} $$ Calculate the sines: $$ \sin 96^\circ \approx 0.9945, \quad \sin 25^\circ \approx 0.4226 $$ 6. **Compute $a$:** $$ a \approx \frac{13 \times 0.9945}{0.4226} \approx \frac{12.9285}{0.4226} \approx 30.59 $$ 7. **Final answer:** The side length opposite the 96° angle is approximately **30.59** units.