In a right triangle, \( \cos (9 x)^{\circ}=\sin (4 x-1)^{\circ} \). Find the smaller of the triangle's two acute angles.

To solve the equation \( \cos (9x)^{\circ} = \sin (4x - 1)^{\circ} \), we can use the identity \( \cos \theta = \sin (90^\circ - \theta) \). This means we can rewrite the equation as: \[ \cos (9x)^{\circ} = \sin (90^\circ - 9x)^{\circ} \] Thus, we have: \[ \sin (90^\circ - 9x)^{\circ} = \sin (4x - 1)^{\circ} \] Using the sine identity, we have two cases to consider: 1. \[ 90^\circ - 9x = 4x - 1 \] 2. \[ 90^\circ - 9x = 180^\circ - (4x - 1) \] ### Solving the first case: First, simplify the equation: \[ 90^\circ - 9x = 4x - 1 \] Adding \(9x\) to both sides gives: \[ 90^\circ = 13x - 1 \] Adding \(1\) to both sides gives: \[ 91^\circ = 13x \] Then divide by \(13\): \[ x = \frac{91^\circ}{13} = 7^\circ \] ### Solving the second case: Now, consider the second case: \[ 90^\circ - 9x = 180^\circ - (4x - 1) \] Simplifying the equation, we have: \[ 90^\circ - 9x = 180^\circ - 4x + 1 \] \[ 90^\circ - 9x = 181^\circ - 4x \] Adding \(9x\) and \(4x\) to both sides brings us to: \[ 90^\circ + 5x = 181^\circ \] \[ 5x = 181^\circ - 90^\circ \] \[ 5x = 91^\circ \] Dividing by \(5\): \[ x = \frac{91^\circ}{5} = 18.2^\circ \] ### Finding the acute angles of the triangle: Now we have two solutions for \(x\): 1. \(x = 7^\circ\) 2. \(x = 18.2^\circ\) The angles of the triangle will thus be \(9x\) and \(4x - 1^\circ\): For \(x = 7^\circ\): - \(9x = 63^\circ\) - \(4x - 1 = 28^\circ - 1 = 27^\circ\) For \(x = 18.2^\circ\): - \(9x = 163.8^\circ\) (not valid, as it exceeds \(90^\circ\)) - \(4x - 1 = 73.8^\circ\) Hence, the acute angles of the triangle for the valid solution \(x = 7^\circ\) are: - \(63^\circ\) and \(27^\circ\). Thus, the smaller of the triangle's two acute angles is: \[ \boxed{27^\circ} \]