# How do you find the angle C formed by the sides of length 2, 3, and 7?

Apr 26, 2015

Not such triangle exists (you can't have a triangle where one side is more that the sum of the lengths of the other two sides).

However, to demonstrate the process:
(and noting that you could get 3 different answers depending upon which side you decide angle C is opposite.)

For demonstration purposes, I will assume the sides are
$a = 2 , b = 3 , \text{ and } c = 7$
with angle C opposite side $c$

The Law of Cosines:
${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cdot \cos \left(C\right)$
can be rearranged as
$\cos \left(C\right) = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}$

$C = \arccos \left(\cos \left(C\right)\right) = \arccos \left(\frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b}\right)$

for the supplied values
$C = \arccos \left(\frac{4 + 9 - 49}{12}\right)$

beyond this you should be able to do the arithmetic and look up the $\arccos \left(\right)$ value in a table or on a calculator... but as noted at the beginning it's all moot since the triangle can't exist.