Question

If $A,B$ where $0°<A,B<180°$and $\sin \left(A\right)+\sin \left(B\right)=\sqrt{\frac{3}{2}}$, $\cos \left(A\right)+\cos \left(B\right)=\sqrt{\frac{1}{2}}$, then what is the value of$(A+B)$?

Answer 1

Step 1: Simplifying the given equations

Given,

1. $\sin \left(A\right)+\sin \left(B\right)=\sqrt{\frac{3}{2}}$
2. $\cos \left(A\right)+\cos \left(B\right)=\sqrt{\frac{1}{2}}$
3. $0°<A,B<180°$

$$\begin{gathered} \Rightarrow \sin A+\sin B=\sqrt{\frac{3}{2}} \\ \Rightarrow 2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=\sqrt{\frac{3}{2}}\dots \dots \dots .\left(i\right)[\mathrm{Using}\mathrm{standard}\mathrm{identity}] \end{gathered}$$

Then,

$$\begin{gathered} \Rightarrow \cos A+\cos B=\frac{1}{\sqrt{2}} \\ \Rightarrow 2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)=\frac{1}{\sqrt{2}}\dots \dots \dots \dots \left(ii\right)[\mathrm{Using}\mathrm{standard}\mathrm{identity}] \end{gathered}$$

Step 2: Find the value

Dividing equation $\left(\mathrm{i}\right)$ by equation $\left(\mathrm{ii}\right)$

$\Rightarrow \frac{\left(2\sin \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right)}{\left(2\cos \left({\displaystyle \frac{A+B}{2}}\right)\cos \left({\displaystyle \frac{A-B}{2}}\right)\right)}=\frac{\sqrt{{\displaystyle \frac{3}{2}}}}{\sqrt{{\displaystyle \frac{1}{2}}}}$

$$\begin{gathered} \Rightarrow \tan \left(\frac{A+B}{2}\right)=\sqrt{3} \\ \Rightarrow \frac{A+B}{2}=60° \\ \Rightarrow A+B=120° \end{gathered}$$

Hence, $A+B=120°$.