Exercise 24 Page 338

We will first $△ABC$ with vertices $A(0,7),B(\text{-}1,\text{-}2),C(2,\text{-}1)$ and $O(0,0).$

In order to show that $m\angle AOB>m\angle AOC,$ we will use the Converse of the Hinge Theorem. We need to show that two sides of $△AOB$ are congruent to two sides of $△AOC.$ Since they share the side $\overline{AO},$ we will find the lengths of other side using the Distance Formula. Let's start with $\overline{OB}.$ Thus, the length of $\overline{OB}$ is $\sqrt{5}.$ We can find the lengths of other sides of the triangles in the same way.

Side	$d=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}$	Length
$\overline{OB}$	$OB=\sqrt{(0-(\text{-}1){)}^2+(0-(\text{-}2){)}^2}$	$OB=\sqrt{5}$
$\overline{OC}$	$OC=\sqrt{(0-2{)}^2+(0-(\text{-}1){)}^2}$	$OC=\sqrt{5}$
$\overline{AB}$	$AB=\sqrt{(0-(\text{-}1){)}^2+(7-(\text{-}2){)}^2}$	$AB=\sqrt{82}$
$\overline{AC}$	$AC=\sqrt{(0-2{)}^2+(7-(\text{-}1){)}^2}$	$AC=\sqrt{68}$

Since two sides of $△AOB$ are congruent to two sides of $△AOC,$ we are able to apply Converse of the Hinge Theorem to compare the included angles.

Converse of the Hinge Theorem

If two sides of one triangle are congruent to two sides of another triangle, and the third sides are not congruent, then the larger included angle is opposite the longer third side.

An illustration of the theorem is given below.

Applying the theorem to the diagram that we draw, we can order the included angles and show that $m\angle AOB>m\angle AOC.$