Question

If in triangle ABC, $a=2,b=1+\sqrt{3}$ and $\angle$C$={60}^o$, then find the value of c.

Answer 1

We are given, $a=2$ & $b=1+\sqrt{3}$ 7 $\angle c={60}^o$

The cosme rule says.

$c^2=a^2+b^2-2ab\cos c$

$\therefore c^2=(2{)}^2+(1+\sqrt{3}{)}^2-2\left(2\right)(1+\sqrt{3})\cos {60}^o$

$\therefore c^2=4+1+2\sqrt{3}+3-(4+4\sqrt{3})\left(\frac{1}{2}\right)$

$\therefore c^2=8+2\sqrt{3}-2-2\sqrt{3}$

$\therefore c^2=6$

$\therefore c=\pm \sqrt{6}$

But, Distance never be negative.

So, $c=\sqrt{6}$