Question

In a triangle $\angle B={45}^0$, side $BC=2(\sqrt{3}+1)$ units and area = $6+2\sqrt{3}$ sq. units. Determine the side b.

Answer 1

$\frac{1}{2}p.BC=2\sqrt{3}(1+\sqrt{3})$
or $p=\frac{4\sqrt{3}(1+\sqrt{3})}{2(\sqrt{3}+1)}=2\sqrt{3}$
or $bsinC=2\sqrt{3}=csinB$ by sine rule ............(1)
$\therefore$ But $\angle B={45}^0$
$ccosB=csinB=2\sqrt{3}$
Again $bcosC+ccosB=a$
$\therefore bcosC=2(\sqrt{3}+2)-2\sqrt{3}=2$ , by (1) ..........(2)
Squaring and adding (1) and (2),
$b^2=4+12=16\therefore b=4$