In Delta ABC,m angle B = 45^{circ} and overli | vedclass

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(C) In $\Delta AMB$,$m \angle M = 90^{\circ}$ and $m \angle B = 45^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,$m \angle MAB = 180

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$5$

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(C) In $\Delta AMB$,$m \angle M = 90^{\circ}$ and $m \angle B = 45^{\circ}$.Since the sum of angles in a triangle is $180^{\circ}$,$m \angle MAB = 180^{\circ} - 90^{\circ} - 45^{\circ} = 45^{\circ}$.Since $m \angle MAB = m \angle B = 45^{\circ}$,$\Delta AMB$ is an isosceles right triangle.Therefore,$AM = BM = 4$.Given $M \in \overline{BC}$,we have $BC = BM + MC$.$7 = 4 + MC$,which implies $MC = 7 - 4 = 3$.In right-angled $\Delta AMC$,by the Pythagorean theorem:$AC^{2} = AM^{2} + MC^{2}$$AC^{2} = 4^{2} + 3^{2} = 16 + 9 = 25$$AC = \sqrt{25} = 5$.

In $\Delta ABC$,$m \angle B = 45^{\circ}$ and $\overline{AM}$ is an altitude,$M \in \overline{BC}$. If $BC = 7$ and $AM = 4$,then $AC = \ldots$

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