Lời giải:
Áp dụng định lý Pitago ta có:
\(\sin A=\frac{BK}{AB}=\frac{BK}{BC}.\frac{BC^2}{BC.AB}=\frac{BK}{KC}.\frac{BK^2+KC^2}{BC.AB}\)
\(=\frac{BK}{KC}.\frac{AB^2-AK^2+KC^2}{BC.AB}=\frac{BK}{KC}.\frac{AC^2-AK^2+KC^2}{BC.AB}\)
\(=\frac{BK}{KC}.\frac{(AK+KC)^2-AK^2+KC^2}{BC.AB}\)
\(=\frac{BK}{KC}.\frac{2KC^2+2AK.KC}{BC.AC}=\frac{BK}{KC}.\frac{2KC.AC}{BC.AC}=2\frac{BK}{KC}.\frac{KC}{BC}\)
\(=2\cos \widehat{KBC}.\sin \widehat{KBC}\)
\(\sin \widehat{KBC}=\sqrt{\frac{2}{3}}\Rightarrow \cos \widehat{KBC}=\sqrt{1-\sin ^2\widehat{KBC}}=\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}\)
Do đó: \(\sin A=2.\frac{1}{\sqrt{3}}.\sqrt{\frac{2}{3}}=\frac{2\sqrt{2}}{3}\)
