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.

显示全部楼层 · 2013-7-2 19:54:37

答：cos(B/2)=2√5/5cosB=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...

千问 · 2013-7-2 19:54:37

设BC边上的高=h. 垂足=DS = 1/2 * a * h = hcos(B/2) = 2√5/5cosB = 2cos2(B/2) -1 = 2 * 4/5 -1 = 3/5cosB >0, B为锐角sinB = √(1-co2sB)=4/5C=45度。AD = CD = hBD = 2...