In a right triangle, \( \sin (x-2)^{\circ}=\cos (5 x+8)^{\circ} \). Find the larger of the triangle's two angles.

To solve the equation \( \sin (x-2)^{\circ}=\cos (5 x+8)^{\circ} \), we can use the identity \( \sin A = \cos (90^{\circ} - A) \). This means we can rewrite the equation as: \[ \sin (x - 2)^{\circ} = \sin (90^{\circ} - (5x + 8)^{\circ}) \] This gives us two possible equations to solve: 1. \( x - 2 = 90 - (5x + 8) \) 2. \( x - 2 = 180 - (90 - (5x + 8)) \) We'll solve the first equation first: \[ x - 2 = 90 - 5x - 8 \] \[ x - 2 = 82 - 5x \] \[ x + 5x = 82 + 2 \] \[ 6x = 84 \] \[ x = 14 \] Now let's solve the second equation: \[ x - 2 = 180 - 90 + 5x + 8 \] \[ x - 2 = 90 + 5x + 8 \] \[ x - 2 = 98 + 5x \] \[ x - 5x = 98 + 2 \] \[ -4x = 100 \] \[ x = -25 \] Since a negative angle does not make sense in the context of triangle angles, we discard \( x = -25 \) and take \( x = 14 \). Now we find the two angles of the triangle: - The angle corresponding to \( x - 2 \) is \( 14 - 2 = 12^{\circ} \). - The angle corresponding to \( 5x + 8 \) is \( 5(14) + 8 = 70 + 8 = 78^{\circ} \). Now we can calculate the third angle of the triangle: \[ 180^{\circ} - (12^{\circ} + 78^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ} \] Clearly, we have angles of \( 12^{\circ} \), \( 78^{\circ} \), and \( 90^{\circ} \). The larger of the two angles in the right triangle is: \[ \boxed{78^{\circ}} \]