a) \(\Delta ABC\) vuông tại C :
\(sinA=sin30^o=\dfrac{BC}{AB}=\dfrac{BC}{10}\)
* \(\rightarrow BC=10.sin30^o=5\left(cm\right)\)
\(AB^2=AC^2+BC^2\) (định lí PItago)
* \(\Rightarrow AC=\sqrt{AB^2-BC^2}=\sqrt{10^2-5^2}=5\sqrt{3}\left(cm\right)\)
Ta có : \(\widehat{ACB}+\widehat{ECB}=180^o\left(kềbù\right)\)
Mà : \(\widehat{ACB}=90^o\rightarrow\widehat{ECB}=180^o-90^o=90^o\)
=> \(\Delta\)BEC vuông tại C
Lại có : \(\widehat{ABC}+\widehat{EBC}=90^o\)(gt)
\(\Rightarrow60^o+\widehat{EBC}=90^o\)
=> \(\widehat{EBC}=30^o\)
\(tanB=tan30^o=\dfrac{CE}{BC}=\dfrac{CE}{5}\)
* \(\Rightarrow CE=5.tan30^o\approx3\left(cm\right)\)
* \(BE=\sqrt{CE^2+BC^2}=\sqrt{3^2+5^2}=\sqrt{34}\left(cm\right)\)
