Question

Tam giác ABC có B-C=20.Kẻ tia phân giác góc A,cắt BC tại D.Số đo góc ADB là

Answer 1

\(\widehat{B}+\widehat{C}=180^0-\widehat{A};\widehat{B}-\widehat{C}=30^0\\ \Rightarrow\widehat{B}=\dfrac{180^0-\widehat{A}+30^0}{2}=\dfrac{210^0-\widehat{A}}{2}=105^0-\dfrac{\widehat{A}}{2}\)

Xét tam giác ABD:

\(\widehat{ADB}=180^0-\widehat{BAD}-\widehat{B}=180^0-\dfrac{\widehat{A}}{2}-\left(105^0-\dfrac{\widehat{A}}{2}\right)\)

\(\Rightarrow\widehat{ADB}=180^0-105^0-\dfrac{\widehat{A}}{2}+\dfrac{\widehat{A}}{2}=75^0\)