4050.11 – Angles in a Special Triangle

Let $\angle A$ be the largest angle; $\angle C$ the smallest.

First Solution. We can use the law of cosines—twice.

$$\begin{aligned}6^2 &= 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cos A \\\cos A &= \dfrac{6^2 - 4^2 - 5^2}{-2 \cdot 4 \cdot 5} = 0.125 \\\angle A &= \arccos(0.125) \approx 82.82°\end{aligned}$$

$$\begin{aligned}4^2 &= 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cos C \\\cos C &= \dfrac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = 0.75 \\\angle C &= \arccos(0.75) \approx 41.41°\end{aligned}$$

Voila!

Second Solution. First we find the area, $\mathcal{A}$, via Heron's formula. Let $s$ be the semi-perimeter of the triangle, i.e., $s = 15/2$. Then,

$$\begin{aligned}\mathcal{A} &= \sqrt{s (s - a)(s - b)(s - c)}, \\&= \sqrt{\dfrac{15}{2} \cdot \dfrac{3}{2} \cdot \dfrac{5}{2} \cdot \dfrac{7}{2}} = \dfrac{15 \sqrt{7}}{4}.\end{aligned}$$

Now we can find the height, $h$, as in figure below. Since

$$\dfrac{15 \sqrt{7}}{4} = \mathcal{A} = \dfrac{bh}{2} = \dfrac{5h}{2}$$

it follows that $h = \dfrac{3 \sqrt 7}{2}$. Now,

$$\begin{aligned}\sin A &= h/4 = \dfrac{3 \sqrt 7}{8} \\\angle A &\approx 82.82° \\\end{aligned}$$

$$\begin{aligned}\sin C &= \dfrac{h}{6} = \dfrac{3 \sqrt 7}{12} \\\angle C &\approx 41.41°\end{aligned}$$

Hey! These solutions are a good advertisement for the law of cosines, don't you agree?