1. **Problem statement:** We have a right triangle ABC with a right angle at C, angle B = 67°, side b = 10.6 opposite to angle C, and side a opposite to angle A. We need to find side a using SOH-CAH-TOA. 2. **Recall SOH-CAH-TOA:** For angle B, sine relates the opposite side to the hypotenuse: $$\sin(B) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c}$$ 3. **Identify sides:** Since angle B = 67°, side a is opposite angle B, side c is hypotenuse. 4. **Use cosine for side b:** $$\cos(B) = \frac{b}{c}$$ so $$c = \frac{b}{\cos(B)}$$ 5. **Calculate hypotenuse c:** $$c = \frac{10.6}{\cos(67^\circ)}$$ 6. **Calculate side a:** Using sine, $$a = c \times \sin(67^\circ) = \frac{10.6}{\cos(67^\circ)} \times \sin(67^\circ)$$ 7. **Simplify expression:** $$a = 10.6 \times \frac{\sin(67^\circ)}{\cos(67^\circ)} = 10.6 \times \tan(67^\circ)$$ 8. **Calculate numeric value:** $$a \approx 10.6 \times 2.35585 = 24.97$$ **Final answer:** $$a \approx 24.97$$