Question

Two sides of a triangle are $2\sqrt{2}$ cm and $2\sqrt{3}$ cm and the angle opposite to the shorter side of the two is $\pi /4.$ The largest possible length of the third side is

• A. $\sqrt{2}(\sqrt{3}+1)cm$
• B. $(6+\sqrt{2})cm$
• C. $(\sqrt{6}-\sqrt{2})cm$
• D. none of these

Answer 1

Answer: (A)

$\sqrt{2}(\sqrt{3}+1)cm$

Let $a=2\sqrt{2}$

And $A={45}^0$

Let $b=2\sqrt{3}$

Therefore, $\cos A=\frac{1}{\sqrt{2}}$

$=\frac{b^2+c^2-a^2}{2bc}$

$=\frac{12+c^2-8}{4\sqrt{3}c}$

$=\frac{4+c^2}{4\sqrt{3}c}$

$=\frac{1}{\sqrt{2}}$

Therefore, $4\sqrt{2}+\sqrt{2}{c}^2=4\sqrt{3}c$

$c^2-2\sqrt{6}c+4=0$

$c=\frac{2\sqrt{6}\pm \sqrt{24-16}}{2}$

$=\sqrt{6}\pm \sqrt{2}$

Therefore, $C=\sqrt{6}-\sqrt{2}$

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

Hence, $C_{max}=\sqrt{6}+\sqrt{2}$

$=\sqrt{2}(\sqrt{3}+1)$