+267 In triangle ABC, cos A=sqrt(7/10) and cos B=sqrt(3/10) Find cos C
+130554 arcos (sqrt (.7)) + arcos (sqrt(.3)) = 90°
cos C = 180 - 90 = cos (90°) = 0
+130554 Assuming a right triangle
Let the legs = AC = sqrt 7 and BC = sqrt 3
The hypotenuse =
sqrt [ (sqrt .7)^2 + (sqrt .3)^2] = sqrt [ 7 + 3] = sqrt 10 = AB
So....noting that cos B = sinA
cos^2 B + cos^2 A =
sin^2 A + cos^2A = 1
(sqrt (3/10)^2 +(sqrt (7/10)^2 = 1
3/10 + 7/10 = 1
Then angle C is opposite the hypotenuse so its measure = 90°
And cos C = cos (90°) = 0
