In a right triangle, \( \sin (9 x+6)^{\circ}=\cos (3 x-4)^{\circ} \). Find the smaller of the triangle's two acute angles.

To solve the equation \( \sin(9x + 6)^{\circ} = \cos(3x - 4)^{\circ} \), we can use the identity \( \sin \theta = \cos(90^\circ - \theta) \). This gives us: \[ 9x + 6 = 90^\circ - (3x - 4) \] Simplifying it, we have: \[ 9x + 6 = 90^\circ - 3x + 4 \] Combining like terms: \[ 9x + 3x = 90^\circ - 4 - 6 \] \[ 12x = 80^\circ \] Dividing both sides by 12: \[ x = \frac{80^\circ}{12} = \frac{20^\circ}{3} \] Now we can find the two acute angles in the triangle. We substitute \( x \) back into the angle expressions. The first acute angle is: \[ \text{Angle 1} = 9x + 6 = 9 \left(\frac{20^\circ}{3}\right) + 6 = 60^\circ + 6^\circ = 66^\circ \] The second acute angle is: \[ \text{Angle 2} = 3x - 4 = 3 \left(\frac{20^\circ}{3}\right) - 4 = 20^\circ - 4^\circ = 16^\circ \] So, the smaller of the two acute angles is: \[ \boxed{16^\circ} \]