Question

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

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

Answer 1

Answer: (C)

$3b$

$3x+5x+4x={180}^o\Rightarrow 12x={180}^o\Rightarrow x={15}^o$

So,

$A=3x={45}^o,B=5x={75}^o,C=4x={60}^o$

$\frac{a}{\sin {45}^o}=\frac{b}{\sin {75}^o}=\frac{c}{\sin {60}^o}$

as $\sin {45}^o=\frac{1}{\sqrt{2}}\sin {75}^o=\frac{\sqrt{3}+1}{2\sqrt{2}}\sin {60}^o=\frac{\sqrt{3}}{2}$

we can write 'a' and 'c' in terms of 'b'

$a=\frac{2b}{\sqrt{3}+1}c=\frac{\sqrt{6}b}{\sqrt{3}+1}$

Now, $a+b+\sqrt{2}c=\frac{2b+\sqrt{3}b+b+b2\sqrt{3}}{\sqrt{3}+1}=\frac{3b(\sqrt{3}+1)}{\sqrt{3}+1}=3b$

Therefore, Correct Answer is $C$