Kẻ đường cao $AD,$ ta có $cosC = CD/b ⇒ b.cosC = b.(CD/b) = CD $Tương tự ta có$: c.cosB = BD ⇒ b.cosC + c.cosB = CD + BD = BC = a $
Kẻ đường cao
A
H" role="presenta
tion" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">AHAH.Ta có
: BH=AB.cos⁡B=c.cos
⁡B" role=
"presentation" style="font-size: 13.696px; display: inline; word-spacing: 0px; position: relative;">BH=AB.cos
B=c.cosBBH=AB.cosB=c.cosBC
H=
AC.
cos⁡C
=b
.cos⁡C" role=
"presentation" style="font-size: 13.696px; display: inline; word-spacing
: 0px; posit
ion: relat
ive;">CH=AC.cosC=b.cosCCH=AC.cosC=b.cosC&a
mp;#x21D2;BH+CH=bc
os⁡C+c.cos
⁡B
" role=
"presentation" style="font-size: 13.696px; word-spacing: 0px; position: relative;">⇒
BH+CH=b
cosC+c.cos
B⇒BH+C
H=bcosC+c.cos
B
hay a=bcos⁡C+
ccos⁡B(1)" role="presentation" style=
"font-size: 13.696px; display: inline; word-spa
cing: 0px; position: relative;">a=bcosC+ccosB(1)