Question

If the angles of triangle $ABC$ are in the ratio $3:5:4$ then find the value of $a+b+c\sqrt{2}$

Answer 1

Given, $A,B,C$ are in ratio

thus $A=$ 3k

$B=$ 5k

$C=$ 4k

adding

$A+B+C$ $=3k+5k+4k$ $={180}^0$
$\therefore 12k={180}^0$
$\therefore k={15}^0$
Hence, $A={45}^0$
$B={75}^0$
$C={60}^0$

by property
Now, $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=n$
Hence,
$a=n\sin A=\frac{n}{\sqrt{2}}$
$b=n\sin B=\frac{n(\sqrt{3}+1)}{2\sqrt{2}}$
$c=n\sin C=\frac{n\left(\sqrt{3}\right)}{2}$

Now, $a+b+\sqrt{2}c$ $=n[\frac{1}{\sqrt{2}}+\frac{\sqrt{3}+1}{2\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}]$
$=n\left[\frac{2+\sqrt{3}+1+2\sqrt{3}}{2\sqrt{2}}\right]$
$=n\left[\frac{3(\sqrt{3}+1)}{2\sqrt{2}}\right]$
$=3nsin{75}^0$
$=3nsinB$
$=3b$