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

We start with the equation \[ \sin (9x+6)^\circ = \cos (3x-4)^\circ. \] Recall the trigonometric identity \[ \cos \theta = \sin (90^\circ-\theta). \] Thus, we can rewrite the right–hand side as \[ \cos (3x-4)^\circ = \sin \Bigl(90^\circ-(3x-4)\Bigr) = \sin (94^\circ-3x). \] This gives \[ \sin (9x+6)^\circ = \sin (94-3x)^\circ. \] When two angles have the same sine, we have \[ 9x+6 = 94-3x + 360k \quad \text{or} \quad 9x+6 = 180-(94-3x)+360k, \] where \(k\) is an integer. Since the two acute angles of a right triangle must add up to \(90^\circ\), our expressions for these angles are \[ A=9x+6 \quad \text{and} \quad B=3x-4. \] Thus, it must be that \[ A+B = 9x+6+3x-4 = 12x+2 = 90. \] Solving for \(x\), \[ 12x = 90-2=88 \quad \implies \quad x = \frac{88}{12} = \frac{22}{3}. \] Now, returning to the sine equation, consider the first possibility with \(k=0\): \[ 9x+6 =94-3x. \] Substitute \(x=\frac{22}{3}\): \[ 9\left(\frac{22}{3}\right)+6 = 66+6=72, \] and \[ 94-3\left(\frac{22}{3}\right) = 94-22 = 72. \] This verifies the equality. Thus, the two acute angles of the right triangle are \[ 9x+6 = 72^\circ \quad \text{and} \quad 3x-4 = 3\left(\frac{22}{3}\right)-4 = 22-4=18^\circ. \] The smaller acute angle is \(\boxed{18^\circ}\).