Question

ABC là một tam giác cân tại A, vẽ \(BK\perp AC\)\(\left(K\in AC\right)\). Nếu \(sin\widehat{KBC}=\frac{\sqrt{2}}{\sqrt{3}}\) vậy \(sin\widehat{BAC}=...\)

Answer 1

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: \(\sin A=2.\frac{1}{\sqrt{3}}.\sqrt{\frac{2}{3}}=\frac{2\sqrt{2}}{3}\)

Answer 2

Hình vẽ: