Bài 4:
Xét tam giác ADB và tam giác ADC có:
\(\left\{{}\begin{matrix}\widehat{ADB}+\widehat{ABC}+\widehat{BAD}=180^0\\\widehat{ADC}+\widehat{ACB}+\widehat{DAC}=180^0\end{matrix}\right.\)(tổng 3 góc trong tam giác)
Mà \(\widehat{BAD}=\widehat{DAC}\)(AD là phân giác)
\(\Rightarrow\widehat{ADB}+\widehat{ABC}=\widehat{ADC}+\widehat{ACB}\)
\(\Rightarrow\widehat{ADC}-\widehat{ADB}=\widehat{ABC}-\widehat{ACB}=20^0\)
Mà \(\widehat{ADC}+\widehat{ADB}=180^0\)(kề bù)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{ADC}=\left(180^0+20^0\right):2=100^0\\\widehat{ADB}=\left(180^0-20^0\right):2=80^0\end{matrix}\right.\)
Bài 5:
Ta có: \(\widehat{BAC}=180^0-\widehat{ABC}-\widehat{ACB}=180^0-30^0-70^0=80^0\)(tổng 3 góc trong tam giác)
Xét tam giác ABC có:
AD là phân giác
\(\Rightarrow\widehat{BAD}=\widehat{DAC}=\dfrac{\widehat{BAC}}{2}=40^0\)
Xét tam giác ABD có:
\(\widehat{ADH}=180^0-\widehat{BAD}-\widehat{ABC}=70^0\)
Xét tam giác AHD vuông tại H:
\(\widehat{HAD}=90^0-\widehat{ADH}=20^0\)
Bài 6:
Ta có: \(5\widehat{A}=3\widehat{B}=15\widehat{C}\)
\(\Rightarrow\dfrac{\widehat{A}}{3}=\dfrac{\widehat{B}}{5}=\dfrac{\widehat{C}}{1}=\dfrac{\widehat{A}+\widehat{B}+\widehat{C}}{3+5+1}=\dfrac{180^0}{9}=20^0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}=60^0\\\widehat{B}=100^0\\\widehat{C}=20^0\end{matrix}\right.\)
Xét tam giác ADB có:
\(\widehat{ADB}=180^0-\widehat{BAD}-\widehat{ABC}\)(tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{ADB}=180^0-\dfrac{1}{2}.60^0-100^0=50^0\)
