In a right triangle, \( \cos (3 x)^{\circ}=\sin (10 x-1)^{\circ} \). Find the larger of the triangle's two acute angles.

To solve the equation \( \cos (3x)^{\circ} = \sin (10x - 1)^{\circ} \), we can use the identity that relates cosine and sine: \[ \cos A = \sin (90^{\circ} - A) \] Applying this identity gives us: \[ \cos(3x)^{\circ} = \sin(90^{\circ} - 3x)^{\circ} \] This allows us to set the right sides equal: \[ \sin(90^{\circ} - 3x)^{\circ} = \sin(10x - 1)^{\circ} \] From the properties of sine, we have two cases: 1. \( 90^{\circ} - 3x = 10x - 1 \) 2. \( 90^{\circ} - 3x = 180^{\circ} - (10x - 1) \) ### Solving the first case: From the first equation, we can rearrange the terms: \[ 90^{\circ} + 1 = 10x + 3x \] \[ 91^{\circ} = 13x \] \[ x = \frac{91^{\circ}}{13} \approx 7^{\circ} \] ### Solving the second case: For the second equation: \[ 90^{\circ} - 3x = 180^{\circ} - 10x + 1 \] \[ 90^{\circ} + 10x - 1 = 180^{\circ} + 3x \] \[ 10x - 3x = 180^{\circ} - 90^{\circ} + 1 \] \[ 7x = 91^{\circ} \] \[ x = 13^{\circ} \] ### Finding the acute angles of the triangle: Now we have two values of \( x \): 1. \( x \approx 7^{\circ} \) 2. \( x \approx 13^{\circ} \) ### Finding the triangle's acute angles: The acute angles of the triangle will be \( x \) and \( 90^{\circ} - x \): - For \( x \approx 7^{\circ} \): - Angles: \( 7^{\circ} \) and \( 90^{\circ} - 7^{\circ} = 83^{\circ} \) - For \( x \approx 13^{\circ} \): - Angles: \( 13^{\circ} \) and \( 90^{\circ} - 13^{\circ} = 77^{\circ} \) Thus, comparing the pairs: - From \( x \approx 7^{\circ} \): \( 7^{\circ} \) and \( 83^{\circ} \) - From \( x \approx 13^{\circ} \): \( 13^{\circ} \) and \( 77^{\circ} \) The larger acute angle from both cases is \( 83^{\circ} \). ### Final answer: The larger of the triangle's two acute angles is \( \boxed{83^{\circ}} \).