Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $A:B:C=3:5:4$, then $[a+b+c\sqrt{2}]$ is equal to?

• A. $2b$
• B. $2c$
• C. $3b$
• D. $3a$

Answer 1

The correct option is C $3b$

Let $\angle A=3x,\angle B=5x,\angle C=4x$

$\Rightarrow 12x={180}^0\Rightarrow x={15}^0\Rightarrow \angle A={45}^0,\angle B={76}^0,\angle C={60}^0$

$\therefore$ By sine formula,

$\frac{a}{\frac{1}{\sqrt{2}}}=\frac{b}{\frac{\sqrt{3}+1}{2\sqrt{2}}}=\frac{c}{\frac{\sqrt{3}}{2}}\Rightarrow \sqrt{2}a=\frac{2\sqrt{2}b}{\sqrt{3}+1}=\frac{2c}{\sqrt{3}}=k$

$\Rightarrow a=\frac{k}{\sqrt{2}},b=\frac{(\sqrt{3}+1)k}{2\sqrt{2}},c=\frac{\sqrt{3}k}{\sqrt{2}}$

$\therefore (a+b+\sqrt{2})=k[\frac{1}{\sqrt{2}}+\frac{(\sqrt{3}+1)}{2\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}]$

$=\frac{k}{2\sqrt{2}}[2+\sqrt{3}+1+2\sqrt{3}]=\frac{k}{2\sqrt{2}}[3+3\sqrt{3}]$

$=\frac{k^3}{2\sqrt{2}}[1+\sqrt{3}]=3b$