Question

Bài 3

Cho tam giác ABC nhọn, có \(\widehat{B}\)\(=70^o\) và \(\widehat{C}\)\(=60^o\). Tia phân giác của góc A cắt BC tại D

Tính\(\widehat{ADB}\) và \(\widehat{ADC}\)

Answer 1

Ta có: \(\widehat{BAC}+\widehat{B}+\widehat{C}=180^0\)

\(\Leftrightarrow\widehat{BAC}=180^0-\widehat{B}-\widehat{C}=180^0-70^0-60^0=50^0\)

\(\Rightarrow\widehat{BAD}=\widehat{CAD}=\dfrac{\widehat{BAC}}{2}=\dfrac{50^0}{2}=25^0\)

Ta có: \(\widehat{ABD}+\widehat{B}+\widehat{ADB}=180^0\)

\(\Leftrightarrow\widehat{ADB}=180^0-\widehat{B}-\widehat{ABD}=180^0-70^0-25^0=85^0\)

Lại có: \(\widehat{ACD}+\widehat{ADB}=180^0\) (2 góc bù nhau)

\(\Leftrightarrow\widehat{ADC}=180^0-\widehat{ADB}=180^0-85^0=95^0\)

Answer 2

ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^o=>\widehat{A}+70^o+60^o=180^o=>\widehat{A}=50^o\)
Ad là  tia phân giác \(=>\widehat{BAD}=\widehat{CAD}=\dfrac{\widehat{A}}{2}=25^o\)
ΔABD có \(\widehat{ADC}=\widehat{B}+\widehat{BAD}=70^o+25^o=95^o\)
\(=>\widehat{ADB}=180^o-\widehat{ADC}=180^o-95^o=85^o\)