Question

Answer 1

XétΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{ACB}+60^0+50^0=180^0\)

=>\(\widehat{ACB}=70^0\)

BD là phân giác của góc ABC

=>\(\widehat{ABD}=\widehat{CBD}=\dfrac{\widehat{ABC}}{2}=\dfrac{50^0}{2}=25^0\)

Xét ΔADB có \(\widehat{CDB}\) là góc ngoài tại đỉnh D

nên \(\widehat{CDB}=\widehat{BAD}+\widehat{ABD}\)

=>\(\widehat{CDB}=25^0+60^0=85^0\)

Ta có: \(\widehat{CDB}+\widehat{ADB}=180^0\)(hai góc kề bù)

=>\(\widehat{ADB}+85^0=180^0\)

=>\(\widehat{ADB}=95^0\)