在三角形ABC已知(6sinA-2sinc)cosB+sin(C-B)=sinA,若b=2根号3,三角形的面积为3根号2,求a c的值
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![在三角形ABC已知(6sinA-2sinc)cosB+sin(C-B)=sinA,若b=2根号3,三角形的面积为3根号2,求a c的值](/uploads/image/z/970736-32-6.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E5%B7%B2%E7%9F%A5%286sinA-2sinc%29cosB%2Bsin%28C-B%29%3DsinA%2C%E8%8B%A5b%3D2%E6%A0%B9%E5%8F%B73%2C%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E9%9D%A2%E7%A7%AF%E4%B8%BA3%E6%A0%B9%E5%8F%B72%2C%E6%B1%82a+c%E7%9A%84%E5%80%BC)
在三角形ABC已知(6sinA-2sinc)cosB+sin(C-B)=sinA,若b=2根号3,三角形的面积为3根号2,求a c的值
在三角形ABC已知(6sinA-2sinc)cosB+sin(C-B)=sinA,若b=2根号3,三角形的面积为3根号2,求a c的值
在三角形ABC已知(6sinA-2sinc)cosB+sin(C-B)=sinA,若b=2根号3,三角形的面积为3根号2,求a c的值
(6sinA - 2sinC)cosB+sin(C-B) = sinA
6sinAcosB - 2sinCcosB +sinCcosB - cosCsinB = sinA
6sinAcosB - (sinCcosB+cosCsinB) = sinA
6sinAcosB - sin(C+B) = sinA
∵sin(C+B) = sin(π - A) = sinA
∴6sinAcosB = 2sinA
cosB = 1/3
sinB = 2√2 /3
S = 1/2 ac sinB = √2 /3 ac = 3√2
∴ac = 9
b² = a²+c² - 2accosB
12 = a²+c² - 2*9*1/3
a²+c² = 18
(a+c)² - 2ac =18
(a+c)² = 18+2ac = 36
a+c = 6
ac=9
∴a=c=3