Question

If the angles of a triangle are ${30}^0$ and ${45}^0$ and the included side is $(\sqrt{3}+1)$ cm, the area of the triangle is $\frac{1}{2}(\sqrt{3}+1)$ sq.cm.

Answer 1

Let $\angle A={30}^0,\angle B={45}^0$ so that $AB=c=(\sqrt{3}+1)$..............(1)
Also $\angle C={180}^0-({30}^0+{45}^0)={105}^0$
$\therefore sinC=sin({90}^0+{15}^0)=cos{15}^0$
or $sinC=cos({45}^0-{30}^0)=\frac{1}{1\sqrt{\left(2\right)}}(\sqrt{3}+1)=\frac{1}{2\sqrt{\left(2\right)}}.c$ by (1) ...........(2)
$\therefore b=sinB.2\sqrt{2}=2\because B={45}^0$
$\therefore △=\frac{1}{2}bcsinA=\frac{1}{2}.(\sqrt{3}+1).\frac{1}{2}=\frac{1}{2}(\sqrt{3}+1)$.