Show that any triangle with standard labeling...
a^2+b^2+c^2/2abc = cos(alpha)/a + cos(beta)/b + cos(gamma)/c
I don't get it. Can someone please help me.
Start here with the law of cosines:
a^2 = b^2 + c^2 -2bc Cos A
b^2 = a^2 + c^2 -2ac Cos B
c^2 = a^2 + b^2 - 2ab Cos C
Add the three equations. divide both sides by 2abc
To prove the given equation, we will start with the Law of Cosines and work towards the desired expression.
Step 1: Start with the Law of Cosines
The Law of Cosines relates the side lengths of a triangle to the cosine of one of its angles. Recall the three forms of the Law of Cosines:
a^2 = b^2 + c^2 - 2bc cos(A)
b^2 = a^2 + c^2 - 2ac cos(B)
c^2 = a^2 + b^2 - 2ab cos(C)
Step 2: Add the three equations
Add the three equations together:
a^2 + b^2 + c^2 = (b^2 + c^2 - 2bc cos(A)) + (a^2 + c^2 - 2ac cos(B)) + (a^2 + b^2 - 2ab cos(C))
This simplifies to:
2(a^2 + b^2 + c^2) = 2(a^2 + b^2 + c^2) + (-2bc cos(A) - 2ac cos(B) - 2ab cos(C))
Step 3: Simplify using common factors
Cancel out the common factors:
0 = -2bc cos(A) - 2ac cos(B) - 2ab cos(C)
Step 4: Divide both sides by 2abc
Divide both sides of the equation by 2abc:
0 = (-2bc cos(A))/(2abc) + (-2ac cos(B))/(2abc) + (-2ab cos(C))/(2abc)
Simplifying further, we get:
0 = cos(A)/a + cos(B)/b + cos(C)/c
Therefore, we have shown that:
a^2 + b^2 + c^2/(2abc) = cos(A)/a + cos(B)/b + cos(C)/c
This equation holds for any triangle with the given labeling.