ແກ້ (AC)^2=(Ab)^2+(BC)^2

$$( A C ) ^ { 2 } = ( A b ) ^ { 2 } + ( B C ) ^ { 2 }$$

ແກ້ສຳລັບ A (complex solution)

$$\left\{\begin{matrix}A=-\left(C^{2}-b^{2}\right)^{-\frac{1}{2}}BC\text{; }A=\left(C^{2}-b^{2}\right)^{-\frac{1}{2}}BC\text{, }&C\neq -b\text{ and }C\neq b\\A\in \mathrm{C}\text{, }&\left(B=0\text{ and }C=b\right)\text{ or }\left(B=0\text{ and }C=-b\right)\text{ or }\left(C=0\text{ and }b=0\right)\end{matrix}\right,$$

ແກ້ສຳລັບ B (complex solution)

$$\left\{\begin{matrix}B=-\sqrt{-\left(\frac{Ab}{C}\right)^{2}+A^{2}}\text{; }B=\sqrt{-\left(\frac{Ab}{C}\right)^{2}+A^{2}}\text{, }&C\neq 0\\B\in \mathrm{C}\text{, }&\left(A=0\text{ or }b=0\right)\text{ and }C=0\end{matrix}\right,$$

ແກ້ສຳລັບ A

$$\left\{\begin{matrix}A=\frac{BC}{\sqrt{C^{2}-b^{2}}}\text{; }A=-\frac{BC}{\sqrt{C^{2}-b^{2}}}\text{, }&|C|>|b|\\A=0\text{, }&\left(C=0\text{ and }b\neq 0\right)\text{ or }\left(B=0\text{ and }|C|\neq |b|\right)\\A\in \mathrm{R}\text{, }&\left(C=0\text{ and }b=0\right)\text{ or }\left(B=0\text{ and }|C|=|b|\right)\end{matrix}\right,$$

ແກ້ສຳລັບ B

$$\left\{\begin{matrix}B=\frac{\sqrt{A^{2}\left(C^{2}-b^{2}\right)}}{|C|}\text{; }B=-\frac{\sqrt{A^{2}\left(C^{2}-b^{2}\right)}}{|C|}\text{, }&\left(A=0\text{ or }|C|\geq |b|\right)\text{ and }C\neq 0\\B\in \mathrm{R}\text{, }&\left(A=0\text{ or }b=0\right)\text{ and }C=0\end{matrix}\right,$$