在三角形ABC中,角A,B,C对应的边为a,b,c.已知
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18、说明:^2——表示平方。
(1) 3cosBcosC+1=3sinBsinC+cos2A
3(cosBcosC-sinBsinC)+1=cos2A
3cos(B+C)+1=2cos^2A-1
3cos(180°-A)=2cos^2A-2
-3cosA=2cos^2A-2
0=2cos^2A+3cosA-2
(2cosA-1)(cosA+2)=0
∵-1=<cosA<=1
∴1=<cosA+2<=3
∴2cosA-1=0
cosA=1/2
A=60°
(2) BC=√7AC
BC^2=AC^2+AB^2-2AC·ABcosA
7AC^2=AC^2+AB^2-2AC·AB·(1/2)
0=AB^2-AC·AB-6AC^2
(AB-3AC)(AB+2AC)=0
∵AB+2AC>0
∴AB-3AC=0
AB=3AC
∵AB=3AD
∴3AD=3AC
AD=AC
∵A=60°
∴ΔACD是等边三角形
AC=AD=CD=5
AB=3AC=15
BC=√7AC=5√7
s=1/2(AB+BC+AC)
=1/2(15+5√7+5)
=10+5√7/2
SΔABC=1/2AB·ACsinA
=1/2×15×5sin60°
=75/2×√3/2
=75√3/4
(1) 3cosBcosC+1=3sinBsinC+cos2A
3(cosBcosC-sinBsinC)+1=cos2A
3cos(B+C)+1=2cos^2A-1
3cos(180°-A)=2cos^2A-2
-3cosA=2cos^2A-2
0=2cos^2A+3cosA-2
(2cosA-1)(cosA+2)=0
∵-1=<cosA<=1
∴1=<cosA+2<=3
∴2cosA-1=0
cosA=1/2
A=60°
(2) BC=√7AC
BC^2=AC^2+AB^2-2AC·ABcosA
7AC^2=AC^2+AB^2-2AC·AB·(1/2)
0=AB^2-AC·AB-6AC^2
(AB-3AC)(AB+2AC)=0
∵AB+2AC>0
∴AB-3AC=0
AB=3AC
∵AB=3AD
∴3AD=3AC
AD=AC
∵A=60°
∴ΔACD是等边三角形
AC=AD=CD=5
AB=3AC=15
BC=√7AC=5√7
s=1/2(AB+BC+AC)
=1/2(15+5√7+5)
=10+5√7/2
SΔABC=1/2AB·ACsinA
=1/2×15×5sin60°
=75/2×√3/2
=75√3/4
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