In a Delta ABC,the ratio of angles A : B : C | vedclass

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(C) Given $A : B : C = 3 : 5 : 4$. Since the sum of angles in a triangle is $180^{\circ}$,we have $3x + 5x + 4x = 180^{\circ}$,which gives $12x = 180^

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$3b$

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(C) Given $A : B : C = 3 : 5 : 4$. Since the sum of angles in a triangle is $180^{\circ}$,we have $3x + 5x + 4x = 180^{\circ}$,which gives $12x = 180^{\circ}$,so $x = 15^{\circ}$.Thus,$A = 45^{\circ}, B = 75^{\circ}, C = 60^{\circ}$.Using the Sine Rule,$\frac{a}{\sin 45^{\circ}} = \frac{b}{\sin 75^{\circ}} = \frac{c}{\sin 60^{\circ}} = K$.Then $a = K \sin 45^{\circ} = \frac{K}{\sqrt{2}}$,$b = K \sin 75^{\circ} = K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$,and $c = K \sin 60^{\circ} = K \frac{\sqrt{3}}{2}$.Now,$a + b + c \sqrt{2} = \frac{K}{\sqrt{2}} + K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right) + K \frac{\sqrt{3}}{2} \cdot \sqrt{2}$.$= \frac{K}{\sqrt{2}} \left( 1 + \frac{\sqrt{3}+1}{2} + \sqrt{3} \right) = \frac{K}{\sqrt{2}} \left( \frac{2 + \sqrt{3} + 1 + 2\sqrt{3}}{2} \right) = \frac{K}{\sqrt{2}} \left( \frac{3 + 3\sqrt{3}}{2} \right) = \frac{3K(\sqrt{3}+1)}{2\sqrt{2}}$.Since $b = K \sin 75^{\circ} = K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$,we have $3b = 3K \left( \frac{\sqrt{3}+1}{2\sqrt{2}} \right)$.Therefore,$a + b + c \sqrt{2} = 3b$.

In a $\Delta ABC$,the ratio of angles $A : B : C = 3 : 5 : 4$. Then $a + b + c \sqrt{2}$ is equal to

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