Question

在三角形ABC中.内角A，B，C的对边为a，b，c，若a=2，C=派/4,cos(B/2)=二倍根号五/5，求三角形ABC的面积S.

Answer 1

答：

cos(B/2)=2√5/5

cosB=2cos²(B/2)-1

=2*(2√5/5)²-1

=3/5

因为：sinB>0，sin²B+cos²B=1

所以：sinB=4/5

所以：

sinA=sin(B+C)=sin(B+π/4)

=(√2/2)(sinB+cosB)

=(√2/2)(7/5)

=7√2/10

根据正弦定理：a/sinA=b/sinB=c/sinC=2R

所以：2/(7√2/10)=c/(√2/2)

解得：c=10/7

所以：

S=(acsinB)/2

=2*(10/7)*(4/5)*(1/2)

=8/7

所以：三角形ABC的面积为8/7

Answer 2

设BC边上的高=h. 垂足=D
S = 1/2 * a * h = h

cos(B/2) = 2√5/5
cosB = 2cos²(B/2) -1 = 2 * 4/5 -1 = 3/5

cosB >0, B为锐角
sinB = √(1-co²sB) =4/5

C=45度。
AD = CD = h
BD = 2-CD = 2-h

AD/BD = h/(2-h) = sinB/cosB = 4/3
3h = 8-4h
7h=8
h= 8/7

所以 S= h= 8/7