Law of Cosines

Wellllllll, here goes another blog... :)

It relates to the lengths of the sides of a triangle to the cosine of one of its angles.

The formula you use is:
c^2=a^2+b^2-2(a)(b)cosangle

Example 1:
angleC=60degrees
side b=5
side a=8

c^2=5^2+8^2-2(5)(8)cos60
*take the square root of everything
side c=7
sin60/7=sinB/5
*cross multiply
5sin60=7sinB
*divide both sides by 7
sinB=.619
*take the inverse
B=sin^-1(.619)
angleB=38.243degrees
*To find angleA, subtract 180-60-38.243
angleA=81.757degrees

Example 2:
angleR=120degrees
side s=14
side t=16

r^2=16^2+14^2-2(16)(14)cos120degrees
*take the square root of everything
side r=26
sin120/26=sinS/14
*cross multiply
14sin120=26sinS
*divide both sides by 26
sinS=.466
*take the inverse
S=sin^-1(.466)
angleS=27.775degrees
*to find angleT, subtract 180-120-27.775
angleT=32.225degrees

Example 3:
side z=41
side y=40
side x=9

40^2=9^2+41^2-2(9)(41)cosY
*Subtract 9^2+41^2 to the other side.
-2(9)(41)cosY=40^2-9^2-41^2
*Divide both sides by -2(9)(41) to get cosY by itself.
cosY=40^2-9^2-41^2/-2(9)(41)
*Take the inverse.
Y=cos^-1 40^2-9^2-41^2/-2(9)(41)
angleY=77.320degrees
41^2=9^2+40^2-2(9)(40)cosZ
*Subtract 9^2+40^2 to the other side.
41^2-9^2-40^2=-2(9)(40)cosZ
*Divide both sides by -2(9)(41) to get cosZ by itself.
41^2-9^2-40^2/-2(9)(40)
*Take the inverse.
Z=cos^-1 41^2-9^2-40^2/-2(9)(40)
angleZ=90degrees
angleX=180-77.320-90=12.68
angleX=12.68degrees