In the opposite figure: \( \overline{\mathrm{BC}} \) is a diameter of the circle M , \( \overline{\mathrm{MD}} \perp \overline{\mathrm{AB}} \) where \( \overline{\mathrm{MD}} \cap \overline{\mathrm{AB}}=\{\mathrm{D}\} \) , \( \mathrm{m}(\angle \mathrm{C})=40^{\circ} \) and \( \mathrm{AC}=8 \mathrm{~cm} \). Find: \( 1 \mathrm{~m}(\angle \mathrm{DMB}) \)

To solve for \( m(\angle DMB) \), we start by recalling that in a circle, the angle subtended by a chord at the center is twice the angle subtended at any point on the circumference. Here, \( \overline{BC} \) is the diameter, so \( m(\angle BAC) = \frac{1}{2} m(\angle BMC) \). Since \( m(\angle C) = 40^\circ \), we find that \( m(\angle BAC) = 40^\circ \) as well. Next, we know that \( m(\angle DMB) \) is equal to \( m(\angle BAC) \) because \( \overline{MD} \) is perpendicular to \( \overline{AB} \). This gives us \( m(\angle DMB) = m(\angle BAC) = 40^\circ \). Therefore, \( m(\angle DMB) = 40^\circ \).