For the acute angles in a right triangle, \( \sin \left(3 x+10^{\circ}\right)=\cos \left(x+24^{\circ}\right) \). What is the value of \( x \) ? (1 point)

To solve the equation \(\sin(3x + 10^\circ) = \cos(x + 24^\circ)\), we can use the identity \(\sin(\theta) = \cos(90^\circ - \theta)\). Therefore, we can write: \[ \sin(3x + 10^\circ) = \cos(x + 24^\circ) \implies 3x + 10^\circ = 90^\circ - (x + 24^\circ) + k \cdot 360^\circ \] Where \(k\) is any integer, but we will first consider \(k = 0\). This simplifies to: \[ 3x + 10^\circ + x + 24^\circ = 90^\circ \] \[ 4x + 34^\circ = 90^\circ \] \[ 4x = 90^\circ - 34^\circ \] \[ 4x = 56^\circ \] \[ x = \frac{56^\circ}{4} = 14^\circ \] Now checking for any possibilities of \(k = 1\) or negative values will lead to angles outside the acute range. Therefore, the only valid solution is: \[ \boxed{14^\circ} \]