在三角形ABC中，a，b，c分别是角A，B，C的对边，角A的余弦值为五分之根号五，角B的正切值为3
\u5728\u4e09\u89d2\u5f62ABC\u4e2d\uff0c\u89d2A B C\u7684\u5bf9\u8fb9\u5206\u522b\u662fa b c\u6b63\u5207\u7b49\u4e8e3\u500d\u6839\u53f77\u6c42C\u7684\u4f59\u5f26\u503ctanC=3\u221a7\uff0c
sinC=3\u221a7cosC\uff0c \uff081\uff09
sinC*sinC+cosC*cosC=1\uff0c \uff082\uff09
\uff081\uff09\u4ee3\u5165\uff082\uff09\u5f0f\u5f97
cosC=1/8\uff0c\u7565\u53bb-1/8\uff0c\u56e0\u4e3aC\u662f\u4e09\u89d2\u5f62\u7684\u5185\u89d2\uff0csinC\u4e3a\u6b63\u503c\uff0c\u6240\u4ee5cosC\u4e5f\u4e3a\u6b63\u3002
1\u3001\u22351+tg²A=1/cos²A\uff0c\u4e14tgA=3\uff0c45\u00b0\uff1cA\uff1c90\u00b0\u2234cos²A=1/10\uff0ccosA=1/\u221a10\uff0csinA=3/\u221a10\uff1b
\u2235cosC=\u221a5/5=1/\u221a5\uff0c\u53ef\u77e545\u00b0\uff1cC\uff1c90\u00b0\u2234sinC=2/\u221a5\uff0c
0\u00b0\uff1cB\uff1c90\u00b0\uff0c\u2235cosB=-cos(A+C)=sinAsinC-cosAcosC=3/\u221a10*2/\u221a5-1/\u221a10*1/\u221a5
=6/\u221a50-1/\u221a50=5/\u221a50=\u221a2/2\uff0c\u2234B=45\u00b0\u3002
2\u3001\u7531b/sinB=c/sinC\uff0c\u5f97b=csinB/sinC\uff0c\u25b3ABC\u7684\u9762\u79efS=(1/2)bcsinA=c²sinBsinA/2sinC
=4²\u00d7\u221a2/2\u00d73/\u221a10\u00f7\uff082\u00d72/\u221a5\uff09=6\u3002
∵cosA=√5/5,A为三角形内角
∴sinA=√[1-(cosA)^2]=2√5/5
∴tanA=sinA/cosA=2
又tanB=3
∴tanC=tan[π-(A+B)]
=-tan(A+B)
=-(tanA+tanB)/(1-tanAtanB)
=-(2+3)/(1-6)
=1
∴C=π/4
(2)
a＝4 ，sinA=2√5/5. C=π/4
根据正弦定理
c/sinC=a/sinA
∴c=asinC/sinA
=4*(√2/2)/(2/√5)
=√10
又tanB=sinB/cosB=3
∴sinB=3cosB代入(sinB)^2+(cosB)^2=1
解得sinB=3/√10
∴SΔABC=1/2acsinB
=1/2*4*√10*3/√10
=6
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