In triangle ABC,if angle A = 45^circ and angl | vedclass

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(D) In $	riangle ABC$,the sum of angles is $180^\circ$. $\angle C = 180^\circ - 45^\circ - 75^\circ = 60^\circ$. Using the Sine Rule,$\frac{a}{\sin A

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$2b$

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(D) In $	riangle ABC$,the sum of angles is $180^\circ$. $\angle C = 180^\circ - 45^\circ - 75^\circ = 60^\circ$. Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$. Thus,$a = k \sin 45^\circ = k \frac{1}{\sqrt{2}}$ and $c = k \sin 60^\circ = k \frac{\sqrt{3}}{2}$. We need to evaluate $a + c\sqrt{2} = k \frac{1}{\sqrt{2}} + k \frac{\sqrt{3}}{2} \sqrt{2} = k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} ight) = k \frac{1 + \sqrt{3}}{\sqrt{2}}$. Since $b = k \sin 75^\circ = k \sin(45^\circ + 30^\circ) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} ight) = k \frac{\sqrt{3} + 1}{2\sqrt{2}}$. Therefore,$2b = 2k \frac{\sqrt{3} + 1}{2\sqrt{2}} = k \frac{\sqrt{3} + 1}{\sqrt{2}}$. Hence,$a + c\sqrt{2} = 2b$.

In triangle $ABC$,if $\angle A = 45^\circ$ and $\angle B = 75^\circ$,then $a + c\sqrt{2} = $

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