ABC中,A=5,B=4,COS(A-B)=31 32
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c应该等于根号7吧,不然此题无解.若c=根号7,则解法如下:∵4sin^2(A+B)/2)-cos2C=7/24sin^2[(180°-C)/2]-cos2C=7/24sin^2(90°-C/2)-c
sin(B+C/2)=sin[B+(π-A-B)/2])=sin[π/2+(B-A)/2]=cos{π/2-[π/2+(B-A)/2]}=cos[(A-B)/2)=4/5cos(A-B)=2cos
由m//n得2sinB*(2cos^2(B/2)-1)=cos2B*(-根号3).2sinBcosB+根号3*cos2B=0.sin2B+根号3*cos2B=0.1/2*sin2B+根号3/2*cos
∵a/sinA=b/sinB∴b*b=4a*a*sinB*sinB化为b^2/(sinB^2)=4a^2a^2/(sinA^2)=4a^2sinA^2=1/4sinA=1/2或sinA=-1/2(舍)
m+n:(cosA+√2-sinA,sinA+cosA)|m+n|=√(x^2+y^2)=√[4+4√2(cosA-sinA)]=2得cosA=sinA,所以A=π/4由余弦定理a^2=b^2+c^2
已知abc成等比数列.sinB=√7/4,b^2=ac,b/a=c/b,根据正弦定理,sinB/sinA=b/a,sinC/sinB=c/b,sinB/sinA=sinC/sinB,(sinB)^2=
3a(cosA)=c(cosB)+b(cosC)即3sinacosa=sinccosb+sinbcocc=sin(b+c)=sina(其实c(cosB)+b(cosC)=a是结论.你画个图作高就出来了
a:b=sinA:sinB得sinB=5/4*sinA=5/8那么cosB=根号39/8以c为底边高=b*sinA=2c=b*cosA+a*cosB=2根号3+5/8根号39则面积=2*c/2=2根号
(1)角ABO=角ACO,角BCO=角CBO,三角形BCO为等腰三角.(2)5个,ef=eb+fc(3)有,beo和cfo;ef=eb+fc
cosC+根号2sinC=根号3cosC=根号3-根号2sinC因为sinC^2+cosC^2=1代入(根号3-根号2sinC)^2+sinC^2=1解得sinC=根号6/3
1.cosC=b2+a2-c2=-2√2
(1)因为m垂直于n,则m*n=0;即sinB*(b*b-a*a-c*c)+(根号3*a*c)*cosB=0;利用余弦定理:a*a+c*c-b*b=2*a*c*cosB;则sinB*cosB*2*a*
题目打错sin(B=C)=应该是sin(B+C)=[否则与正弦定理矛盾]1∵sinA=sin(B+C)=2sinB,∴a=2b=2√52cosB=13/(6√5)[余弦定理]sinB=√11/(6√5
呃...可能是110°吧!A为40,所以ABC+ACB为180-40=140,又因为两个角分线,所以OBC+OCB=140/2=70,所以BOC=180-70=110
因为B<C,则B小于90度,所以cosB=3/5所以sin(A+C)=4/5,cos(A+C)=-3/5而cos(2A+C)=cos(A+(A+C))=cosAcos(A+C)-sinAsin(A+C
由题意得sinA,sinB,sinC也成等比数列,所以sinB的平方等于sinA*sinC=(5/13)*(5/13)=25/169.cotA+cotC=(cosA*sinC+cosC*sinA)/s
解,1,向量m⊥向量n∴m*n=0∴b*(cosA-2cosC)+(a-2c)*cosB=0利用正弦定理,b=sinB*2Rc=sinC*2R∴sinB*(cosA-2cosC)+(sinA-2sin
A=90度a=2c^2/aa^2=2c^2a=根号2c所以a=根号2bc=b等腰直角三角形
(1)证明:连接AO,则AO⊥PA,∠AOC=2∠B=120°,∴∠AOP=60°,∴∠P=30°,又∵OA=OC,∴∠ACP=30°,∴∠P=∠ACP,∴AP=AC.(2)在Rt△PAO中,∠P=3
cosB=(a^2+c^2-b^2)/2aca.b.c成等比数列b^2=acc=2a代入得到cosB=(a^2+c^2-ac)/2ac=(a^2+4a^2-2a^2)/(4a^2)=3/4