a) Xét \(\bigtriangleup ABC\), có:
\(\widehat{A} + \widehat{B} + \widehat{C} = 180^{\circ}\)
\(\widehat{A} = 180^{\circ} - (\widehat{B} + \widehat{C})\)
\(\widehat{A}\)= \(180^{\circ} - (70^{\circ} + 30^{\circ}) = 80^{\circ}\)
Hay: \(\widehat{BAC}= 80^{\circ}\)
b) Ta có: \(\widehat{DAC} = \frac{\widehat{BAC}}{2}=\frac{80^{\circ}}{2}= 40^{\circ}\) (AD là tia phân giác của \(\widehat{BAC}\))
Xét \(\bigtriangleup ADC\), có:
\(\widehat{DAC} + \widehat{ADC} + \widehat{C}= 180^{\circ}\)
\(40^{\circ} + \widehat{ADC} + 30^{\circ} = 180^{\circ}\)
\(70^{\circ} + \widehat{ADC}= 180^{\circ}\)
\(=> \widehat{ADC}= 180^{\circ} - 70^{\circ} = 110^{\circ}\)
Mà: \(\widehat{ADH} + \widehat{ADC} = 180^{\circ} (kb)\)
Nên: \(\widehat{ADH}= 180^{\circ} - \widehat{ADC}= 180^{\circ} - 110^{\circ}= 70^{\circ}\)
c) Xét \(\bigtriangleup AHD\) vuông tại H, ta có:
\(\widehat{HAD}= 90^{\circ} - \widehat{ADH}= 90 - 70^{\circ}= 20^{\circ}\)(2 góc phụ nhau)
