Question

If two sides of a triangle are $2 \sqrt { 3 }$ and $2 \sqrt { 2 }$ the angle

opposite the shorter side is $45 ^ { \circ }$. The maximum value of the

third side is

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

Answer 1

Answer: (B)

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

Answer 2

Let $a = 2 \sqrt { 3 } , b = 2 \sqrt { 2 }$ $\therefore B = 45 ^ { \circ }$

$\therefore$ $\frac { a } { \sin A } = \frac { b } { \sin B } = \frac { c } { \sin C }$ ⇒ $\frac { 2 \sqrt { 3 } } { \sin A } = \frac { 2 \sqrt { 2 } } { \sin 45 ^ { \circ } } = \frac { c } { \sin C }$ ......(i)

$\therefore$ $\sin A = \frac { \sqrt { 3 } } { 2 }$ ⇒ $A = 60 ^ { \circ }$.

$\therefore$ $C = 180 ^ { \circ } - A - B = 75 ^ { \circ }$

From (i) $c = 4 \sin C = 4 \sin \left( 45 ^ { \circ } + 30 ^ { \circ } \right)$ = $\sqrt { 2 } + \sqrt { 6 }$.