Question

Answer 1

Xét ΔABC có \(\widehat{BAC}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{BAC}+20^0+40^0=180^0\)

=>\(\widehat{BAC}=120^0\)

AD nằm giữa AB và AC

=>\(\widehat{BAD}+\widehat{CAD}=\widehat{BAC}\)

=>\(2\cdot\widehat{BAD}+\widehat{BAD}=120^0\)

=>\(\widehat{BAD}=\dfrac{120^0}{3}=40^0\)

=>\(\widehat{CAD}=2\cdot40^0=80^0\)

Xét ΔCDA có \(\widehat{CDA}+\widehat{ACD}+\widehat{CAD}=180^0\)

=>\(\widehat{CDA}+80^0+40^0=180^0\)

=>\(\widehat{CDA}=60^0\)