1. Given \(\cos A = \cos B\) with \(A,B\) acute (0° < A,B < 90°).\
2. Subtract the two sides: \(\cos A - \cos B = 0\).\
3. Use the trigonometric identity\
$$\cos A - \cos B = -2\sin\frac{A+B}{2}\,\sin\frac{A-B}{2}.$$\
Hence\
$$-2\sin\frac{A+B}{2}\,\sin\frac{A-B}{2}=0.$$\
4. For acute angles, \(0°
0\).\
5. Since the product is zero and \(\sin\frac{A+B}{2}
eq0\), we must have\
$$\sin\frac{A-B}{2}=0.$$\
6. The sine of an angle is zero only when the angle is an integer multiple of \(180°\). Because \(\frac{A-B}{2}\) lies between \(-45°\) and \(45°\) (as A and B are acute), the only possible multiple is 0°; thus\
$$\frac{A-B}{2}=0° \quad\Rightarrow\quad A-B=0°.$$\
7. Hence \(A = B\).\
8. Therefore, if \(\cos A = \cos B\) for acute angles, the angles must be equal.
Answer: \(\displaystyle \angle A = \angle B\).