Question

If two sides of triangle are $\sqrt{12}$ and $\sqrt{8}$, and the angle opposite the shorter side is ${45}^o$, the maximum value of the third side can be?

• A. $\sqrt{2}$
• B. $\sqrt{6}$
• C. $\sqrt{6}-\sqrt{2}$
• D. $\sqrt{6}+\sqrt{2}$

Answer 1

Answer: (D)

$\sqrt{6}+\sqrt{2}$

The shorter side is $\sqrt{8}$.
Let the length of the unknown side be $x$.
Hence,
$\cos {45}^0=\frac{12+x^2-8}{2x\sqrt{12}}$
$\frac{1}{\sqrt{2}}=\frac{x^2+4}{4\sqrt{3}.x}$
$4\sqrt{3}.x=\sqrt{2}.x^2+4\sqrt{2}$
$x^2-2\sqrt{6}x+4=0$
$x=\frac{2\sqrt{6}\pm \sqrt{24-16}}{2}$
$=\sqrt{6}\pm \sqrt{2}$.
Hence maximum value of the unknown side will be
$=\sqrt{6}+\sqrt{2}$.