$ABC,A^{\prime }{B}^{\prime }{C}^{\prime }$ are two triangles such that $A+A^{\prime }={180}^0,B=B^{\prime }$ Then

The correct option is A $a{a}^{\prime }=b{b}^{\prime }+c{c}^{\prime }$

$A+B+C={180}^o=A^{\prime }+B^{\prime }+C^{\prime }$

We have,

$\sin A^{\prime }=\sin ({180}^o-A)=\sin A$ and $\cos A^{\prime }=\cos ({180}^o-A)=-\cos A$

Similarly, $\sin B^{\prime }=\sin B$ and $\cos B^{\prime }=\cos B$

${\sin }^2{A}^{\prime }{\sin }^2{B}^{\prime }={\sin }^{2}A{\sin }^{2}B$

$\Rightarrow {\sin }^{2}A(1-{\cos }^{2}B)={\sin }^{2}B(1-{\cos }^{2}A)$

$\Rightarrow {\sin }^{2}A={\sin }^{2}B+{\sin }^{2}A{\cos }^{2}B-{\cos }^{2}A{\sin }^{2}B$

$\Rightarrow \sin A\sin A^{\prime }={\sin }^{2}B+(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)$

$\Rightarrow \sin A\sin A^{\prime }=\sin B\sin B^{\prime }+\left(\sin \right(A+B\left)\right)\left(\sin \right(A-B\left)\right)$

$\Rightarrow \sin A\sin A^{\prime }=\sin B\sin B^{\prime }+\left(\sin C\right)\sin ({180}^o-A^{\prime }-B^{\prime })$

$\Rightarrow \sin A\sin A^{\prime }=\sin B\sin B^{\prime }+\sin C\sin C^{\prime }$

Applying sine rule we get,

$a{a}^{\prime }=b{b}^{\prime }+c{c}^{\prime }$