Answer:
129.8 approximately
Step-by-step explanation:
So this sounds like a problem for the Law of Cosines. The largest angle is always opposite the largest side in a triangle.
So 11 is the largest side so the angle opposite to it is what we are trying to find. Let's call that angle, X.
My math is case sensitive.
X is the angle opposite to the side x.
Law of cosines formula is:
[tex]x^2=a^2+b^2-2ab \cos(X)[/tex]
So we are looking for X.
We know x=11, a=4, and b=8 (it didn't matter if you called b=4 and a=8).
[tex]11^2=4^2+8^2-2(4)(8)\cos(X)[/tex]
[tex]121=16+64-64\cos(X)[/tex]
[tex]121=80-64\cos(X)[/tex]
Subtract 80 on both sides:
[tex]121-80=-64\cos(X)[/tex]
[tex]41=-64\cos(X)[/tex]
Divide both sides by -64:
[tex]\frac{41}{-64}=\cos(X)[/tex]
Now do the inverse of cosine of both sides or just arccos( )
[these are same thing]
[tex]\arccos(\frac{-41}{64})=X[/tex]
Time for the calculator:
X=129.8 approximately
