1. **Problem statement:** We have a right triangle with vertices H, I, and J, where side HI is perpendicular to side IJ, forming a right angle at vertex I. Angle H measures 30 degrees, and side IJ (opposite angle H) is 16 units long. We need to find the value of $\sin J$ rounded to the nearest hundredth. 2. **Relevant formula:** In a right triangle, the sine of an angle is defined as the ratio of the length of the side opposite the angle to the hypotenuse: $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$$ 3. **Identify sides:** Since angle I is the right angle, the hypotenuse is side HJ. Side IJ is adjacent to angle J and opposite to angle H. 4. **Find the hypotenuse using angle H:** Using angle H = 30 degrees, side IJ = 16 is opposite angle H. So, $$\sin 30^\circ = \frac{IJ}{HJ} = \frac{16}{HJ}$$ Since $\sin 30^\circ = 0.5$, we have: $$0.5 = \frac{16}{HJ}$$ 5. **Solve for hypotenuse HJ:** $$HJ = \frac{16}{0.5} = 32$$ 6. **Find side HI (adjacent to angle H):** Using cosine of angle H, $$\cos 30^\circ = \frac{HI}{HJ}$$ $$HI = HJ \times \cos 30^\circ = 32 \times \frac{\sqrt{3}}{2} = 16\sqrt{3}$$ 7. **Find $\sin J$:** Angle J is complementary to angle H (since angles H and J add to 90 degrees in a right triangle). So, $$\sin J = \cos H = \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.87$$ **Final answer:** $$\sin J \approx 0.87$$