Question

三角形a=2，c=3.cosB=1/4，求b的值和sinC的值

Answer 1

解法一：利用余弦定理：
b^2=a^2+c^2-2ac*cosB
=4+9-2*2*3*(1/4)
=13-3
=10
b=10^(1/2)
sinB=(1-cos^2B)^(1/2)
=15^(1/2)/4
利用正弦定理：
b/sinB=c/sinC
sinC=c*sinB/b
=[3*15^(1/2)/4}/10^(1/2)
=3*6^(1/2)/8
解法二：利用勾股定理和简单的三角函数定义求
过A点作AD垂直BC于D点
在Rt△ADB中,
BD=AB*cosB=c*cosB=3/4
AD=AB*sinB=AB*(1-cos^2B)^(1/2)=(3/4)*15^(1/2)
DC=BDC-BD=a-BD=2-3/4=5/4
在Rt△ADC中,AC=b=(AD^2+DC^2)^(1/2)
b=[(9*15/16)+25/16]^(1/2)
=(160/16)^(1/2)
故,b=根号10
sinC=AD/AC=AD/b=(3/4)*(15/10)^(1/2)
故,sinC=(3/8)*根号6