在三角形ABC中,角ABC所对的边分别是abc已知a=3,cosA=（根号6）/3,B=A+（π/2）.1.求b值2.求ABC

Question

在三角形ABC中,角ABC所对的边分别是abc
已知a=3,cosA=（根号6）/3,B=A+（π/2）.1.求b值 2.求ABC

Answer 1

（1） 由于 B = A + π/2
所以 A一定为锐角,
、 所以 sinA = √（1 - sin^2 A ) = √ [ 1 - (√6/3)^2 = √(1/3) = √3/3
由正弦定理有：a/sinA = b/sinB
而sinB = sin(A +π/2 ) = cosA = √6 / 3
因此,b = asinB/sinA = 3 × （√6/3) / ( √3/3)
= 3√2
(2) 由 B = A + π/2知,B为钝角：
所以 cosB = -√(1 - sin^2 B) = -√[ 1 - (√6/3)^2] = - √3/3
由于A + B +C = π
所以sinC = sin(A + B) = sinAcosB + cosAsinB = √3/3 × (-√3/3) + √6/3 × √6/3 = - 1/3 + 2/3 = 1/3
所以三角形ABC的面积S = 1/2 absinC = 1/2 × 3 × 3√2 × 1/3 = 3√2/2