In a Delta ABC, if A:B:C = 3:5:4, then a + b | vedclass

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(C) Given $A:B:C = 3:5:4.$ Let $A = 3x, B = 5x, C = 4x.$Since $A + B + C = 180^{\circ},$ we have $12x = 180^{\circ},$ so $x = 15^{\circ}.$Thus,$A = 45

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$3b$

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(C) Given $A:B:C = 3:5:4.$ Let $A = 3x, B = 5x, C = 4x.$Since $A + B + C = 180^{\circ},$ we have $12x = 180^{\circ},$ 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}},$$b = K \sin 75^{\circ} = K \sin(45^{\circ} + 30^{\circ}) = K(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2}) = \frac{K(\sqrt{3} + 1)}{2\sqrt{2}},$$c = K \sin 60^{\circ} = \frac{K\sqrt{3}}{2}.$Now,$a + b + c\sqrt{2} = \frac{K}{\sqrt{2}} + \frac{K(\sqrt{3} + 1)}{2\sqrt{2}} + \frac{K\sqrt{3}}{2} \cdot \sqrt{2} = \frac{K}{\sqrt{2}} + \frac{K\sqrt{3} + K}{2\sqrt{2}} + \frac{K\sqrt{3}}{\sqrt{2}} = \frac{2K + K\sqrt{3} + K + 2K\sqrt{3}}{2\sqrt{2}} = \frac{3K + 3K\sqrt{3}}{2\sqrt{2}} = \frac{3K(1 + \sqrt{3})}{2\sqrt{2}}.$Since $3b = 3 \cdot \frac{K(\sqrt{3} + 1)}{2\sqrt{2}},$ we see that $a + b + c\sqrt{2} = 3b.$

In a $\Delta ABC,$ if $A:B:C = 3:5:4,$ then $a + b + c\sqrt{2}$ is equal to

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