MCQ
In a $\triangle A B C, A: B: C=3: 5: 4$. Then $a+b+c \sqrt{2}$ is equal to
  • A
    2b
  • B
    2c
  • 3b
  • D
    3a

Answer

Correct option: C.
3b
(C) Let $x$ be the common multiple.
$\begin{array}{ll}\therefore & A + B + C =12 x=180^{\circ} \Rightarrow x=15^{\circ} \\ \therefore & A =45^{\circ}, B =75^{\circ}, C =60^{\circ}\end{array}$
$\frac{ a }{\sin 45^{\circ}}=\frac{ b }{\sin 75^{\circ}}=\frac{ c }{\sin 60^{\circ}}= k$
$\therefore \quad a=\frac{1}{\sqrt{2}} k, b=\frac{\sqrt{3}+1}{2 \sqrt{2}} k, c=\frac{\sqrt{3}}{2} k$
$\therefore \quad a+b+c \sqrt{2}=\frac{3+3 \sqrt{3}}{2 \sqrt{2}}=3 b$

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