In a right triangle, $\sin (4 x-5)^{\circ}=\cos (3 x+4)^{\circ}$. Find the smaller of the triangle's two acute angles. Answer Attempt 1 out of 2

Step 1 ：Given that in a right triangle, $\sin (4 x-5)^{\circ}=\cos (3 x+4)^{\circ}$, we know that the sine of an angle in a right triangle is equal to the cosine of its complement. Therefore, the two angles given in the equation must be complementary.

Step 2 ：This means that their sum is equal to 90 degrees. We can set up the equation as follows: (4x - 5) + (3x + 4) = 90.

Step 3 ：Solving this equation gives us the value of x as 13.

Step 4 ：Substituting x = 13 back into the expressions for the angles, we get the two angles as 47 degrees and 43 degrees.

Step 5 ：Comparing these two angles, we find that the smaller angle is 43 degrees.

Step 6 ：Final Answer: The smaller of the triangle's two acute angles is $\boxed{43}$ degrees.