Question

Cho △ABC, tia pg AD ( D ∈ BC ) Tính \(\widehat{ADB}\) và \(\widehat{ADC}\). Biết \(\widehat{B}-\widehat{C}=30^o\)

Answer 1

\(\widehat{BAD}+\widehat{ABD}+\widehat{ADB}=\widehat{CAD}+\widehat{ACD}+\widehat{ADC}\)

mà \(\widehat{BAD}=\widehat{CAD}\)

nên \(\widehat{ABD}+\widehat{ADB}=\widehat{ACD}+\widehat{ADC}\)

=>\(\widehat{ADB}+30^0+\widehat{C}=\widehat{ADC}+\widehat{C}\)

\(\Leftrightarrow\widehat{ADB}-\widehat{ADC}=-30^0\)

mà \(\widehat{ADB}+\widehat{ADC}=180^0\)

nên \(\widehat{ADB}=\dfrac{180^0-30^0}{2}=75^0\)

=>\(\widehat{ADC}=105^0\)