Expert: Sherman D. - 11/1/2008

Question
use the vectors to find the cosines of the interior angles of the triangle with vertices (0,1)(1,-2) (4,1)

Answer
(0,1) and (1,-2)
D = sqrt((1 - 0)^2 + (-2 - 1)^2)
D = sqrt(1 + 9)
D = sqrt(10)

(0,1) and (4,1)
D = sqrt((4 - 0)^2 + (1 - 1)^2)
D = sqrt(16)
D = 4

(1,-2) and (4,1)
D = sqrt((4 - 1)^2 + (1 - (-2))^2)
D = sqrt(9 + 9)
D = sqrt(18)
D = 3sqrt(2)

using that

sqrt(10)^2 = sqrt(16)^2 + sqrt(18)^2 - 2(sqrt(16)sqrt(18))cos(A)
10 = 16 + 18 - 2(12sqrt(2))cos(A)
10 = 34 - 24sqrt(2)cos(A)
-24 = -24sqrt(2)cos(A)
sqrt(2)/2 = cos(A)
A = 45

sqrt(16)^2 = sqrt(10)^2 + sqrt(18)^2 - 2(sqrt(18)sqrt(10))cos(B)
16 = 10 + 18 - 2(6sqrt(5))cos(B)
16 = 28 - 12sqrt(5)cos(B)
-12 = -12sqrt(5)cos(B)
cos(B) = sqrt(5)/5
B = 63.435

sqrt(18)^2 = sqrt(10)^2 + sqrt(16)^2 - 2(sqrt(10)sqrt(16))cos(C)
18 = 10 + 16 - 2(4sqrt(10))cos(C)
18 = 26 - 8sqrt(10)cos(C)
-8 = -8sqrt(10)cos(C)
cos(C) = sqrt(10)/10
C = 71.565

The angles are

A = 45°
B = 63°26'5.82"
C = 71°33'54.18"

their cosines are

cos(A) = sqrt(2)/2
cos(B) = sqrt(5)/5
cos(C) = sqrt(10)/10

if your wondering, i used the law of cosines to solve along with the distance formula.