Question

cho\(\Delta\)ABC , tia phân giác \(\widehat{A}\) cắt ABC tại D. tính \(\widehat{ADC}\) biết rằng:

a)\(\widehat{B}=70^o;\widehat{c}=30^o\)

b)\(\widehat{B}-\widehat{c}=40^o\)

các bn giúp mk nhé

Answer 1

a: \(\widehat{BAC}=180^0-70^0-30^0=80^0\)

=>\(\widehat{CAD}=40^0\)

\(\widehat{ADC}=180^0-40^0-30^0=110^0\)

b: \(\widehat{B}-\widehat{C}=40^0\)

nên \(\widehat{B}=\widehat{C}+40^0\)

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

\(\Leftrightarrow\widehat{C}+40^0+\widehat{ADB}=\widehat{C}+\widehat{ADC}\)

\(\Leftrightarrow\widehat{ADB}-\widehat{ADC}=-40^0\)

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

nên \(-2\cdot\widehat{ADC}=\dfrac{-40^0-180^0}{2}=-110^0\)

hay \(\widehat{ADC}=55^0\)