a) \(\Delta ABC\) có: \(\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o\)
\(\widehat{BAC}=180^o-\left(\widehat{ABC}+\widehat{ACB}\right)\)
\(\widehat{BAC}=180^o-100^o\)
\(\widehat{BAC}=80^o\)
b) Ta có: \(\widehat{BAD}=\widehat{CAD}=\widehat{\dfrac{BAC}{2}}=\dfrac{80^o}{2}=40^o\)
\(\Delta ABD\) có: \(\widehat{BAD}+\widehat{ABD}+\widehat{ADB}=180^o\)
\(\widehat{ADB}=180^o-\left(\widehat{BAD}+\widehat{ABD}\right)\)
\(\widehat{ADB}=180^o-110^o\)
\(\widehat{ADB}=70^o\)
c) \(\Delta AHD\) vuông tại H, ta có:
\(\widehat{HAD}+\widehat{ADH}=90^o\)
\(\widehat{HAD}=90^o-\widehat{ADH}\)
\(\widehat{HAD}=90^o-70^o\)
\(\widehat{HAD}=20^o\).
a) ΔABCΔABC có: ˆBAC+ˆABC+ˆACB=180oBAC^+ABC^+ACB^=180o
ˆBAC=180o−(ˆABC+ˆACB)BAC^=180o−(ABC^+ACB^)
ˆBAC=180o−100oBAC^=180o−100o
ˆBAC=80oBAC^=80o
b) Ta có: ˆBAD=ˆCAD=ˆBAC2=80o2=40oBAD^=CAD^=BAC2^=80o2=40o
ΔABDΔABD có: ˆBAD+ˆABD+ˆADB=180oBAD^+ABD^+ADB^=180o
ˆADB=180o−(ˆBAD+ˆABD)ADB^=180o−(BAD^+ABD^)
ˆADB=180o−110oADB^=180o−110o
ˆADB=70oADB^=70o
c) ΔAHDΔAHD vuông tại H, ta có:
ˆHAD+ˆADH=90oHAD^+ADH^=90o
ˆHAD=90o−ˆADHHAD^=90o−ADH^
ˆHAD=90o−70oHAD^=90o−70o
ˆHAD=20oHAD^=20o.
