Question

在三角形中，ABC所对边分别为abc且满足余弦A/2=2根号5/5，c*b*余弦A=3，求三角形面积，若b+c=6，求a值

Answer 1

(1)
三角形面积 = 1/2 * b*c * sin(A)
cos(A/2) = 2sqrt(5)/5 => sin(A/2) = sqrt(5)/5
那么sin(A) = sin(2 * A/2) = 2sin(A/2) * cos(A/2) = 4/5
cos(A) = sqrt(1-(sin(A))^2) = 3/5

b*c * cos(A) = 3
b*c = 5
三角形面积 = 1/2 * b*c * sin(A) = 1/2 * 5 * 4/5 = 2
(2)
b+c = 6; b*c = 5 => (b,c) = (1,5)或(5,1)
由余弦定理知:
a^2 = b^2 + c^2 - 2*b*c*cos(A)
b^2 + c^2 = (b+c)^2 - 2b*c = 36 - 2* 10 = 16
a^2 = 16 - 2*5 *3/5 = 10
a = sqrt(10)