Question

In a triangle ABC , two sides are $\sqrt{12}$ and $\sqrt{8}$, the angle opposite to the shortest side is ${45}^0$, then the third side is

• A. $2\sqrt{2}-\sqrt{6}$
• B. $\sqrt{2}+\sqrt{6}$
• C. $\sqrt{6}-\sqrt{2}$
• D. None of these

Answer 1

The correct option is C $\sqrt{6}-\sqrt{2}$

Let the third side be $c$

$a=\sqrt{12}$

$b=\sqrt{8}$

$\angle B={45}^{\circ }$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\frac{1}{\sqrt{2}}=\frac{12+c^2-8}{2c\sqrt{12}}$

$c\sqrt{24}=c^2+4$

squaring on both sides,

$24c^2=c^4+8c^2+16$

$c^4-16c^2+16=0$

solving the above equation, we get $c=\sqrt{6}+\sqrt{2}$ or $c=\sqrt{6}-\sqrt{2}$

but since shorter so negative sign is considered