Question

If the angles of a triangle are ${30}^o$ and ${45}^o$ and the included side is $(\sqrt{3}+1)$, then what is the area of the triangle?

• A. $\frac{\sqrt{3}+1}{2}$
• B. $2(\sqrt{3}+1)$
• C. $\frac{\sqrt{3}+1}{3}$
• D. $\frac{\sqrt{3}-1}{2}$

Answer 1

The correct option is A $\frac{\sqrt{3}+1}{2}$

$a=\sqrt{3}+1,B={30}^{\circ },C={40}^{\circ }A+B+C={180}^{\circ }\Rightarrow C=105^{\circ }\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}b=\frac{a\sin B}{\sin A}b=\frac{(\sqrt{3}+1)\sin {30}^{\circ }}{\sin {105}^{\circ }}b=\sqrt{2}\Delta =\frac{1}{2}ab\sin C\Delta =\frac{1}{2}(\sqrt{3}+1)\sqrt{2}\sin {105}^{\circ }\Delta =\frac{\sqrt{3}+1}{2}$