Chapter 8.6, Problem 52PPS

To determine

The largest angle in triangle ABC.

Answer to Problem 52PPS

  $m\angle A=82.1$

Explanation of Solution

Given:

  $\begin{array}{l}A=\left(-3,6\right)\\ B=\left(4,2\right)\\ C=\left(-5,1\right)\end{array}$

Calculation:

The length of the sides will be:

  $\begin{array}{l}AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ AB=\sqrt{{\left(4-\left(-3\right)\right)}^2+{\left(2-6\right)}^2}\\ AB=\sqrt{49+16}\\ AB=\sqrt{65}\\ \\ BC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ BC=\sqrt{{\left(-5-4\right)}^2+{\left(1-2\right)}^2}\\ BC=\sqrt{81+1}\\ BC=\sqrt{82}\end{array}$

  $\begin{array}{l}AC=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ AC=\sqrt{{\left(-5-\left(-3\right)\right)}^2+{\left(1-6\right)}^2}\\ AC=\sqrt{4+25}\\ AC=\sqrt{29}\end{array}$

The largest angle in triangle ABC will be:

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ \cos A=\frac{{\left(AB\right)}^2+{\left(AC\right)}^2-{\left(BC\right)}^2}{2\left(AB\right)\left(AC\right)}\\ \cos A=0.1367\\ m\angle A={\cos }^{-1}0.1367\\ m\angle A=82.1\end{array}$