在△ABC中,C-A=二分之π,sinB=三分子一求∶①sinA的值②设AC=√6,求散△ABC的面积

Question

在△ABC中,C-A=二分之π,sinB=三分子一求∶①sinA的值 ②设AC=√6,求散△ABC的面积

Answer 1

C – A =π/2 => C = A +π/2 => sinC = sin(A +π/2) = cosA 以及 cosC = cos(A +π/2) = -sinA ,（也就是说C是钝角,A是锐角）1）由于B = π- C – A ,所以 sinB = sin(π- C – A) = sin(A + C) = sinAcosC + cosAsinC = -sin2A + cos2A = cos2A = 1 – 2sin2A = 1/3 => sin2A = 1/3 => sinA = √3/3 ；2）由正弦定理,a/sinA = b/sinB => a = b(sinA)/sinB ,由于A 是锐角,所以 cosA = √(1 - sin2A) = √6/3 = sinC => SΔABC = (1/2)absinC = (1/2)b2sinAsinC/sinB = (1/2) * (√6)2 *(√3/3) * (√6/3) / (1/3) = 3√2 .