\(\widehat{ADC}=180^0-80^0=100^0\)
Ta có: \(\widehat{ADB}+\widehat{B}+\widehat{BAD}=\widehat{ADC}+\widehat{C}+\widehat{CAD}\)
mà \(\widehat{BAD}=\widehat{CAD}\)
nên \(\widehat{ADB}+\widehat{B}=\widehat{ADC}+\widehat{C}\)
=>\(\widehat{B}-\widehat{C}=100^0-80^0=20^0\)
=>\(\dfrac{3}{2}\widehat{C}-\widehat{C}=20^0\)
=>\(\widehat{C}=40^0\)
\(\widehat{B}=\dfrac{3}{2}\cdot40^0=60^0\)
\(\widehat{BAC}=180^0-40^0-60^0=80^0\)
Đúng 0
Bình luận (0)
