Chapter 5, Problem 25CUR

To determine

To find: $\angle 1$ , $\angle 2$ , $\angle 3$ and $\angle 4$

Answer to Problem 25CUR

  $\angle 1=k°$

  $\angle 2=\left(45-k\right)°$

  $\angle 3=\left(45-k\right)°$

  $\angle 4=k°$

Explanation of Solution

Given:

The diagram:

  $m\angle VOZ=90$

$\overline{OW}$ is an altitude of $△VOZ$ .

$\overline{OX}$ bisects $\angle VOZ$ .

$\overline{OY}$ is median of $△VOZ$ .

  $m\angle Z=k$

Concept Used:

Theorem 5-15: In right triangle, the midpoint of the hypotenuse is equidistant from the three vertices.

Theorem 4-1: Opposite angles of equal sides in triangle are equal.

Theorem 3-11: The addition of all angles of a triangle is 180.

Consider the figure shown below:

Consider the right triangle $△VOZ$ .

Now, $\overline{VZ}$ is the hypotenuse of the right triangle $△VOZ$ and $\overline{OY}$ is median of $△VOZ$ .

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

Thus,

  $OY=YZ$

Now, consider the triangle $△OYZ$ , here $\overline{OY}\cong \overline{YZ}$ , so by theorem 4-1, the angles $\angle YOZ$ and $\angle Z$ are congruent.

That is,

  $m\angle YOZ=m\angle Z$

$m\angle 4=k$ .... (1)

Now it is given that $\overline{OX}$ bisects $\angle VOZ$ , so

  $m\angle VOX=m\angle XOZ$

Also, $m\angle VOZ=90$ , so

  $\begin{array}{c}m\angle VOZ=90\\ m\angle VOX+m\angle XOZ=90\\ m\angle VOX+m\angle VOX=90\end{array}$

  $\begin{array}{c}2m\angle VOX=90\\ m\angle VOX=\frac{90}{2}\\ m\angle VOX=45\end{array}$

Since, $m\angle VOX=m\angle XOZ$

So,

  $m\angle XOZ=m\angle VOX=45$

Now consider,

  $\begin{array}{c}m\angle XOZ=45\\ m\angle XOY+m\angle YOZ=45\\ m\angle 3+m\angle 4=45\end{array}$

  $\begin{array}{c}m\angle 3+k=45\\ m\angle 3=45-k\end{array}$

$m\angle 3=15$ .... (2)

Apply theorem 3-11 on triangle $△VOZ$ ,

  $\begin{array}{c}m\angle V+m\angle VOZ+m\angle Z=180\\ m\angle V+90+k=180\\ m\angle V+k=180-90\end{array}$

  $\begin{array}{c}m\angle V+k=90\\ m\angle V=90-k\end{array}$

Now, it is given that $\overline{OW}$ is an altitude of $△VOZ$ .

So $m\angle VWO=90$

Now, apply theorem 3-11 on triangle $△VWO$ ,

  $\begin{array}{c}m\angle V+m\angle VWO+m\angle VOW=180\\ 90-k+90+m\angle 1=180\\ m\angle 1+180-k=180\\ m\angle 1=180-180+k\end{array}$

  $m\angle 1=k$ .... (3)

Now consider,

  $\begin{array}{c}m\angle VOX=45\\ m\angle VOW+m\angle WOX=45\\ m\angle 1+m\angle 2=45\\ k+m\angle 2=45\end{array}$

  $m\angle 2=45-k$ .... (4)