Question

在△ABC中，C-A=π/2，sinB=1/3 （1）求sinA的值 （2）设AC=根号6，求△ABC的面积

Answer 1

C-A=π/2
所以,C=A+π/2，
所以sinC=cosA,cosC=-sinA,
sinB=sin(A+C)=sinAcosC+cosAsinC
=-sin^A+cos^A
=1-2sin^A=1/3,
1/3=sin^A,sinA>0,
所以sinA=√3/3.
（2）AC=√6，由正弦定理，
BC=AC*sinA/sinB=3√2，
sinC=cosA=√6/3，
△ABC的面积=（1/2）BC*AC*sinC
=(1/2)*3√2*√6*√6/3
=3√2.

Answer 2

sinA=sin(B+C) = sin(pi/2+B+A) = cos(B+A)
sinA = cosB cosA -sinAsinB = cosA sqrt(8)/3 - sinA/3
sinA /cos A = sqrt(8)/4
sinA = 1/sqrt(3)

b = sqrt(6)
sinB/b = sinA/a
a = b sinA/sinB = sqrt(6)/sqrt(3) * 3 = 3sqrt(2)
sinC = sin(pi/2+A) = cosA = sqrt(6)/3
面积=1/2absinC = 3sqrt(12)*sqrt(6)/6 = 3sqrt(2)