Cosine value of a / 2 + C = 4 / 5, area of triangle ABC = vector AB * vector BC, find cosine value of A Who will help solve the problem of grade one in senior high school!

Cosine value of a / 2 + C = 4 / 5, area of triangle ABC = vector AB * vector BC, find cosine value of A Who will help solve the problem of grade one in senior high school!

Triangle ABC area = * = | ab | * | BC | cos (180-b)=
According to the triangle area formula
Area = 1 / 2 ×| ab | * | BC | SINB
So SINB / 2 = cos (180-b)
We get tanb = - 2
And because a + B + C = 180
tan（A+C）=2
Cos (A / 2 + C) = 4 / 5, so tan (A / 2 + C) = 3 / 4
So tan (A / 2) = Tan (a + C - (A / 2 + C)) = 1 / 2
According to the universal formula, cosa = 3 / 5