a: Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)
=>\(\widehat{BAC}=180^0-70^0-30^0=80^0\)
b: AD là phân giác của góc BAC
=>\(\widehat{BAD}=\widehat{CAD}=\dfrac{\widehat{BAC}}{2}=\dfrac{80^0}{2}=40^0\)
Xét ΔABD có \(\widehat{ABD}+\widehat{ADB}+\widehat{BAD}=180^0\)
=>\(\widehat{ADB}+40^0+70^0=180^0\)
=>\(\widehat{ADB}=180^0-110^0=70^0\)
b: ΔAHD vuông tại H
=>\(\widehat{HAD}+\widehat{HDA}=90^0\)
=>\(\widehat{HAD}=90^0-70^0=20^0\)
`a)`Xét `ΔABC` có :
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)(tổng 3 góc trong 1 Δ)
`=> `\(\widehat{A}=180^0-70^0-30^0=80^0\)
`b)` vì `AD` là phân giác \(\widehat{A}\)
`=>` \(\widehat{BAD}=\widehat{DBC}=\dfrac{1}{2}\widehat{A}=\dfrac{1}{2}80^0=40^0\)
Xét `ΔABD` có :
\(\widehat{BAD}+\widehat{ADB}+\widehat{B}=180^0\)
`=>`\(\widehat{ADB}=180^0-40^0-70^0=70^0\)
Có :`AH⊥BC`
`=>`\(\widehat{AHD}=90^0\)
Xét `ΔAHD` có :
\(\widehat{HAD}+\widehat{AHD}+\widehat{ADH}=180^0\)
`=>` \(\widehat{HAD}=180^0-90^0-70^0=20^0\)
