The Law of Cosines
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Key Questions

Cosider the triangle:
(Picture source: Wikipedia)you can relate the sides of this triangle in a kind of "extended" form of Pitagora's Theorem giving:
#a^2=b^2+c^22bc*cos(alpha)#
#b^2=a^2+c^22ac*cos(beta)#
#c^2=a^2+b^22ab*cos(gamma)# As you can see you use this law when your triangle is not a rightangled one.
Example:
Consider the above triangle in which:
#a=8 cm#
#c=10 cm#
#beta=60Â°# therefore:
#b^2=a^2+c^22ac*cos(beta)#
#b^2=8^2+10^22*8*10*cos(60Â°)# but#cos(60Â°)=1/2#
so:#b^2=84 and b=sqrt(84)= 9,2 cm# 
The Law of Cosine can only be used if two sides and their enclosed angle are known.
In all other cases the Law of Sines has to be used.

Yes, the Law of Cosines works for all triangles.
However, the proof depends on the shape a triangle, more precisely, how an altitude from some vertex falls onto the opposite side.
For example, consider a triangle
#Delta ABC# with vertices#A# ,#B# and#C# , corresponding angles#alpha# ,#beta# and#gamma# and correspondingly opposite sides#a# ,#b# and#c# .
Let's prove the Law of Cosines that states:
#a^2+b^22*a*b*cos(gamma) = c^2# Let's draw altitude
#AH# from vertex#A# to an opposite side#BC# with an intersection of this altitude and a side#BC# at point#H# .
There are different cases of a location of point#H# relatively to vertices#B# and#C# .
It can lie in between vertices#B# and#C# .
It can lie outside of#BC# on a continuation of this side beyond vertex#B# or beyond vertex#C# .Assume that a base of this altitude, point
#H# , is lying on the continuation of#BC# beyond a point#C# (so,#C# is in between#B# and#H# ) and prove the Law of Cosines in this case. Other cases are similar to this one.Let's use the following symbols for segments involved:
#AH# is#h#
#BH# is#a_1#
#CH# is#a_2#
Then, since#C# lies in between#B# and#H# ,
#a = a_1  a_2# or#a_1=a+a_2#
Since both#Delta ABH# and#Delta ACH# are right triangles, by trigonometric dependency between hypotenuse, catheti and angles and by Pythagorean Theorem
#h = b*sin(pigamma) = b*sin(gamma)#
#a_2 = b*cos(pigamma) = b*cos(gamma)#
#c^2 = h^2+a_1^2 = b^2*sin^2(gamma)+(ab*cos(y))^2 =#
#=b^2*sin^2(gamma)+a^22*a*b*cos(gamma)+b^2*cos^2(gamma)=#
#=a^2+b^2(sin^2(gamma)+cos^2(gamma))2*a*b*cos(gamma)=#
#=a^2+b^22*a*b*cos(gamma)# End of Proof
When point
#H# lies in between vertices#B# and#C# or on a continuation of side#BC# beyond vertex B, the proof is similar.See Unizor Trigonometry  Simple Identities  Law of Cosines for visual presentation and more detailed information.

Find C using the Law of Cosines.
#c^2=b^2+a^22abCos(C)#
This can also be written as:
#a^2=b^2+c^22bcCosA#
#b^2=a^2+c^22acCosB# Say you have a triangle with Angles A, B, and C and sides a, b, and c as shown below.
In order to find the angle, the problem will most likely give you the length of the 3 sides.Just for explanation purposes, let's give these side lengths values.
a=9, c=8, and b=5.If we substituted all of these values into our Law of Cosines equation, the new equation would look like:
#8^2=5^2+9^22(5)(9)Cos(C)# You wold then simplify the equation so that it comes down to
42=90Cos(C)Isolate the Cos(C):
Cos(C)=42/90You now have to take the inverse of Cosine to isolate Angle C.
#C=Cos^1(42/90)#
Round to however your teacher wants you to, but you should get about 62.18 degrees.
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