* Tính \(\widehat{yBC}\)
Do BA là tia phân giác của \(\widehat{yBC}\)
\(\Rightarrow\widehat{yBC}=2.\widehat{ABC}=2.50^0=100^0\)
* Tính \(\widehat{KAB}\)
Ta có: \(\widehat{KAB}=\widehat{BAC}=30^0\)
* Tính \(\widehat{BKA}\)
Do \(\widehat{KBC}\) và \(\widehat{yBC}\) là hai góc kề bù
\(\Rightarrow\widehat{KBC}+\widehat{yBC}=180^0\)
\(\Rightarrow\widehat{KBC}=180^0-\widehat{yBC}=180^0-100^0=80^0\)
\(\Rightarrow\widehat{KBA}=\widehat{KBC}+\widehat{ABC}=80^0+50^0=130^0\)
\(\Delta KBA\) có:
\(\widehat{BKA}+\widehat{KAB}+\widehat{KBA}=180^0\) (tổng ba góc trong \(\Delta KBA\))
\(\Rightarrow\widehat{BKA}=180^0-\left(\widehat{KAB}+\widehat{KBA}\right)=180^0-\left(30^0+130^0\right)=20^0\)
