Inverse Functions: Arccosine and SSS (Part 20)

The Law of Cosines also recognizes when the purported sides of a triangle are impossible.

Solve $\triangle ABC if$latex a = 16\$, $b = 20$, and $c = 40$.

Hopefully students would recognize that $c > a + b$, thus quickly demonstrating that the triangle is impossible. However, this also falls out of the Law of Cosines:

$c^2 = a^2 + b^2 - 2 a b \cos \gamma$

$1600 = 256 + 400 - 640 \cos \gamma$

$944 =-640 \cos \gamma$

$-1.475 = \cos \gamma$

Since the cosine of an angle can’t be less than -1, we can conclude that this is impossible.

Stated another way, we have the implications (since $a$, $b$, and $c$ are all positive)

$c > a + b \Longleftrightarrow c^2 > (a+b)^2$

$\Longleftrightarrow a^2 + b^2 - 2 a b \cos \gamma > a^2 + 2 a b + b^2$

$\Longleftrightarrow -2 a b \cos \gamma > 2 a b$

$\Longleftrightarrow \cos \gamma < -1$

Since the last statement is impossible, so is the first one.