1. **State the problem:** We need to find the sine of angle $H$ in a right triangle $IHG$ where $\angle G$ is the right angle. 2. **Recall the sine definition:** In a right triangle, $\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}$. 3. **Identify sides relative to $\angle H$:** - Opposite side to $\angle H$ is $IG = 60$. - Hypotenuse is $IH = 68$. 4. **Apply the sine formula:** $$\sin(H) = \frac{IG}{IH} = \frac{60}{68}$$ 5. **Simplify the fraction:** $$\frac{60}{68} = \frac{\cancel{4} \times 15}{\cancel{4} \times 17} = \frac{15}{17}$$ 6. **Final answer:** $$\sin(H) = \frac{15}{17}$$ This is a proper simplified fraction representing the sine of angle $H$.