1. **State the problem:** We have a right triangle where $\sin(7x+2)^\circ = \cos(6x-3)^\circ$. We need to find the larger of the two acute angles. 2. **Recall the identity:** For any angle $\theta$, $\sin \theta = \cos(90^\circ - \theta)$. 3. **Apply the identity:** Since $\sin(7x+2)^\circ = \cos(6x-3)^\circ$, it implies: $$7x + 2 = 90 - (6x - 3)$$ 4. **Solve the equation:** $$7x + 2 = 90 - 6x + 3$$ $$7x + 2 = 93 - 6x$$ Add $6x$ to both sides: $$7x + 6x + 2 = 93$$ $$13x + 2 = 93$$ Subtract 2 from both sides: $$13x = 91$$ Divide both sides by 13: $$x = \frac{91}{13}$$ 5. **Calculate the angles:** $7x + 2 = 7 \times \frac{91}{13} + 2 = \frac{637}{13} + 2 = \frac{637}{13} + \frac{26}{13} = \frac{663}{13} \approx 51$ degrees $6x - 3 = 6 \times \frac{91}{13} - 3 = \frac{546}{13} - 3 = \frac{546}{13} - \frac{39}{13} = \frac{507}{13} \approx 39$ degrees 6. **Identify the larger acute angle:** The two acute angles are approximately 51 degrees and 39 degrees. The larger is $51^\circ$. **Final answer:** The larger acute angle is $51^\circ$.