Question

Cho ΔABC có \(\widehat{B}\) - \(\widehat{C}\) = 30^o. Tia phân giác của \(\widehat{A}\) cắt BC ở D. Tính số đo \(\widehat{ADB}\)

Answer 1

Xét \(\Delta ADB\), có: \(\widehat{ADB}+\widehat{BAD}+\widehat{B}=180^o\)

\(\Rightarrow\widehat{ADB}=180^o-\dfrac{1}{2}\widehat{BAC}-\widehat{B}\)

\(=180^o-\dfrac{1}{2}\left(180^o-\widehat{B}-\widehat{C}\right)-\widehat{B}\)

\(=180^o-\dfrac{1}{2}\left(180^o-\widehat{B}-\widehat{B}+30^o\right)-\widehat{B}\)

\(=180^o-\dfrac{1}{2}\left(210^o-2\widehat{B}\right)-\widehat{B}\)

\(=180^o-105^o=75^o\)