In a right triangle, \( \sin (7 x+2)^{\circ}=\cos (6 x-3)^{\circ} \). Find the larger of the triangle's two acute angles.

We start with the equation given by the right triangle: \[ \sin(7x + 2)^{\circ} = \cos(6x - 3)^{\circ} \] Using the complementary angle identity, we can write: \[ \sin(θ) = \cos(90^{\circ} - θ) \] This means we have: \[ 7x + 2 = 90^{\circ} - (6x - 3) \] Now we simplify the right side: \[ 7x + 2 = 90^{\circ} - 6x + 3 \] Combining like terms results in: \[ 7x + 2 = 93^{\circ} - 6x \] Next, we isolate \(x\): \[ 7x + 6x = 93^{\circ} - 2 \] \[ 13x = 91^{\circ} \] \[ x = \frac{91^{\circ}}{13} = 7^{\circ} \] With \(x = 7^{\circ}\), we can substitute this value back into the expressions for the angles. First, we find \(7x + 2\): \[ 7x + 2 = 7(7^{\circ}) + 2 = 49^{\circ} + 2^{\circ} = 51^{\circ} \] Next, we find \(6x - 3\): \[ 6x - 3 = 6(7^{\circ}) - 3 = 42^{\circ} - 3^{\circ} = 39^{\circ} \] Now we have the acute angles of the triangle: - Angle 1 (from the sine): \(51^{\circ}\) - Angle 2 (from the cosine): \(39^{\circ}\) To find the larger of the two acute angles, we compare: \[ 51^{\circ} > 39^{\circ} \] Thus, the larger of the triangle's two acute angles is \[ \boxed{51^{\circ}}. \]