1. **Problem (c):** Find the leg $y$ of a right triangle with hypotenuse 13 cm and angle 25° opposite to $y$. 2. In a right triangle, the sine of an angle gives the ratio of the opposite leg to the hypotenuse. Here, $\sin 25^\circ = \frac{y}{13}$. 3. Solve for $y$: $$ y = 13 \times \sin 25^\circ $$ 4. Calculate $\sin 25^\circ \approx 0.4226$, so $$ y \approx 13 \times 0.4226 = 5.494 $$ 5. Therefore, $y \approx 5.49$ cm. 6. **Problem (d):** Find leg $z$ of a right triangle where one leg is 14 cm, and the angle opposite this leg is 29°. The right angle is between legs $z$ and 14 cm. 7. Because 29° is opposite the 14 cm leg, $14 = z \times \tan 29^\circ$ as $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$. Here, $14$ is opposite the 29° angle, so $$ \tan 29^\circ = \frac{14}{z} \implies z = \frac{14}{\tan 29^\circ} $$ 8. Calculate $\tan 29^\circ \approx 0.5543$, so $$ z \approx \frac{14}{0.5543} = 25.26 $$ 9. Hence, $z \approx 25.26$ cm.