Question

In a triangle ABC , if $\angle A={45}^0,\angle C={60}^0,$ the a+c is equal to

• A. $\left(\frac{\sqrt{2}+\sqrt{2}}{2}\right)k.$
• B. $\left(\frac{\sqrt{2}+\sqrt{3}}{2}\right)k.$
• C. $\left(\frac{\sqrt{2}+\sqrt{3}}{3}\right)k.$
• D. $\left(\frac{\sqrt{3}+\sqrt{3}}{2}\right)k.$

Answer 1

The correct option is B $\left(\frac{\sqrt{2}+\sqrt{3}}{2}\right)k.$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin c}=k$ $[A=45°,B=75°,C=60°]$

$\Rightarrow a=k\sin A=k\sin 45°=\frac{k}{\sqrt{2}}$

$c=k\sin C=k\sin 60°=\frac{\sqrt{3}}{2}$

$b=k\sin B=k\sin 75°$

$\Rightarrow a+c=k(\frac{1}{\sqrt{3}}+\frac{\sqrt{3}}{2})=\left(\frac{\sqrt{2}+\sqrt{3}}{2}\right)k$

Hence, the answer is $\left(\frac{\sqrt{2}+\sqrt{3}}{2}\right)k.$