Question

Two of a sides of a triangle are of lengths $\sqrt{6}$ and 4 and the angle opposite to smaller side is ${30}^0$. How many such triangles are possible? Find the length of their third side and area.

Answer 1

$\frac{4}{sinC}=\frac{\sqrt{6}}{sin{30}^0}\therefore sinC=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}}$
There will be two values of C which will be supplementary. Hence there will be two such triangles.
Again by cosine rule,
$(\sqrt{6}{)}^2=a^2+16-2a.4cos{30}^0$
or $a^2-4\sqrt{3}a+10=0$
$\therefore a=\frac{4\sqrt{3}\pm \sqrt{48-40}}{2}=2\sqrt{3}\pm \sqrt{2}$
Above gives the values of two sides.