Tính góc BAC biết tam giác ABC có góc B =70 độ, góc C=30 độ

a) Xét \(\Delta ABC\) có: \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^o\) Mà \(\widehat{B}=70^o\); \(\widehat{C}=30^o\)
\(\Rightarrow70+30+\widehat{BAC}=180^o\)
\(\Rightarrow\widehat{BAC}=180-70-30=80^o\)
Vậy \(\widehat{BAC}=80^o\)

b) Vì AD là tia phân giác của \(\widehat{BAC}\):

\(\Rightarrow\widehat{BAD}=\widehat{DAC}=\dfrac{\widehat{BAC}}{2}=\dfrac{80}{2}=40^o\)

Xét \(\Delta ADC\) có: \(\widehat{DAC}=\widehat{ACD}+\widehat{ADC}=180^o\)

Mà \(\widehat{BAD}=40^o\); \(\widehat{ACD}=30^o\)

\(\Rightarrow40+30+\widehat{ADC}=180^o\)

\(\Rightarrow\widehat{ADC}+180-40-30=110^o\)

Vì \(\widehat{ADC}\) kề bù với góc \(\widehat{ADH}\):

\(\Rightarrow\widehat{ADC}+\widehat{ADH}=180^o\)

Mà \(\widehat{ADC}=110^o\)

\(\Rightarrow110+\widehat{ADH}=180^o\)

\(\Rightarrow\widehat{ADH}=180-110=70^o\)

Vậy \(\widehat{ADH}=70^o\)

c) Vì \(AH\perp BC\Rightarrow\widehat{AHD}=90^o\)

Xét \(\Delta AHD\) có: \(\widehat{AHD}+\widehat{ADH}+\widehat{HAD}=180^o\)

Mà \(\widehat{AHD}=90^o\); \(\widehat{ADH}=70^o\)

\(\Rightarrow90+70+\widehat{HAD}=180^o\)

\(\Rightarrow\widehat{HAD}=180-90-70=20^o\)

Vậy \(\widehat{HAD}=70^o\)