In a right triangle, \( \cos (6 x)^{\circ}=\sin (5 x-4)^{\circ} \). Solve for \( x \). Round your answer to the nearest hundredth if necessary.

To solve the equation \( \cos (6 x)^{\circ} = \sin (5 x - 4)^{\circ} \), we can use the identity \( \sin \theta = \cos (90^\circ - \theta) \). Thus, we can rewrite the equation as: \[ \cos (6x)^{\circ} = \cos \left(90^\circ - (5x - 4)\right)^{\circ} \] This implies: \[ 6x = 90^\circ - (5x - 4) + k \cdot 360^\circ \quad \text{or} \quad 6x = -(90^\circ - (5x - 4)) + k \cdot 360^\circ \] Let's first solve the equation: \[ 6x = 90^\circ - 5x + 4 + k \cdot 360^\circ \] This simplifies to: \[ 6x + 5x = 94^\circ + k \cdot 360^\circ \] \[ 11x = 94^\circ + k \cdot 360^\circ \] \[ x = \frac{94^\circ + k \cdot 360^\circ}{11} \] For \( k = 0 \): \[ x = \frac{94^\circ}{11} \approx 8.545^\circ \] Next, solve the second equation: \[ 6x = -(90^\circ - 5x + 4) + k \cdot 360^\circ \] This simplifies to: \[ 6x = -90^\circ + 5x - 4 + k \cdot 360^\circ \] \[ 6x - 5x = -94^\circ + k \cdot 360^\circ \] \[ x = -94^\circ + k \cdot 360^\circ \] For \( k = 0 \): \[ x = -94^\circ \, \text{(discard, since we want a positive angle)} \] Now let's check \( k = 1 \): \[ x = -94^\circ + 360^\circ = 266^\circ \, \text{(not valid in a triangle)} \] Thus, the only valid solution for \( x \) is approximately: \[ \boxed{8.55}^\circ \]