Chapter 5.5, Problem 6ST2

To determine

To find: $AM$ and $m\angle ACM$

Answer to Problem 6ST2

  $AM=17$

  $m\angle ACM=67$

Explanation of Solution

Given:

The diagram:

M is the midpoint of the hypotenuse $\overline{AB}$

Consider the figure shown below:

Consider the triangle $△ACB$ .

Now, $\overline{AB}$ is the hypotenuse of the right triangle $△ACB$ and M is its midpoint.

So, by theorem 5-15, the midpoint M of the hypotenuse $\overline{AB}$ of a right triangle $△ACB$ is equidistant from the three vertices.

Thus,

  $BM=AM=CM=17$

Now, consider the triangle $△MCB$ , here $\overline{BM}\cong \overline{AM}$ , so by theorem 4-1, the angles $\angle MCB$ and $\angle MBC$ are congruent.

That is,

  $m\angle MCB=m\angle MBC=23$

Now it is given that $m\angle ACB=90$ , so

  $\begin{array}{c}m\angle ACB=90\\ m\angle ACM+m\angle MCB=90\\ m\angle ACM+23=90\end{array}$

  $\begin{array}{c}m\angle ACM=90-23\\ m\angle ACM=67\end{array}$