1. **Problem statement:** We have a right triangle with an altitude $d$ drawn to the hypotenuse $c$, dividing it into segments $p$ and $q$. The sides are labeled $a$, $b$, and the angle opposite side $a$ is $\alpha$. We need to find three different ratios of segment lengths equal to $\sin \alpha$. 2. **Recall the definition of sine:** $$\sin \alpha = \frac{\text{opposite side}}{\text{hypotenuse}}$$ In the original triangle, the side opposite $\alpha$ is $a$, and the hypotenuse is $c$, so: $$\sin \alpha = \frac{a}{c}$$ 3. **Use similar triangles:** The altitude $d$ creates two smaller right triangles inside the big triangle, each similar to the original triangle and to each other. - Triangle with sides $a$, $d$, $p$ (adjacent to $\alpha$) - Triangle with sides $b$, $d$, $q$ 4. **First ratio:** In the original triangle: $$\sin \alpha = \frac{a}{c}$$ 5. **Second ratio:** Consider the smaller triangle with hypotenuse $p$ and side $a$ opposite $\alpha$: $$\sin \alpha = \frac{a}{p}$$ But since $p$ is part of the hypotenuse in the smaller triangle, this ratio also equals $\sin \alpha$. 6. **Third ratio:** Consider the smaller triangle with hypotenuse $b$ and side $d$ opposite $\alpha$: $$\sin \alpha = \frac{d}{b}$$ **Summary of ratios equal to $\sin \alpha$:** $$\sin \alpha = \frac{a}{c} = \frac{a}{p} = \frac{d}{b}$$ **Final answer:** - $\sin \alpha = \frac{a}{c}$ - $\sin \alpha = \frac{a}{p}$ - $\sin \alpha = \frac{d}{b}$