Chapter 3.4, Problem 23WE

(a)

To determine

To Find: The measure of $\angle ABD$ .

Answer to Problem 23WE

  $m\angle ABD=22°$

Explanation of Solution

Given Information: In the figure shown below, $\overline{AB}\perp \overline{BC}\text{ and }\overline{BD}\perp \overline{AC}$ .

  

Also, $m\angle C=22°$ .

Calculation:

In $\Delta ABC$ , the sum of all interior angles is equal to 180 degrees.

  $\begin{array}{l}m\angle CAB+m\angle CBA+m\angle BCA=180°\\ m\angle CAB+90°+22°=180°\left[\because \overline{AB}\perp \overline{BC}\right]\\ m\angle CAB=180°-90°-22°\\ m\angle CAB=68°\end{array}$

In $\Delta ABD$ , the sum of all interior angles is equal to 180 degrees.

  $\begin{array}{l}m\angle BAD+m\angle ADB+m\angle ABD=180°\\ 68°+90°+m\angle ABD=180°\left[\because \overline{BD}\perp \overline{AC}\right]\\ m\angle ABD=180°-90°-68°\\ m\angle ABD=22°\end{array}$

Therefore, the measure of $\angle ABD$ is 22°.

(b)

To determine

To Find: The measure of $\angle ABD$ .

Answer to Problem 23WE

  $m\angle ABD=23°$

Explanation of Solution

Given Information: In the figure shown below, $\overline{AB}\perp \overline{BC}\text{ and }\overline{BD}\perp \overline{AC}$ .

  

Also, $m\angle C=23°$ .

Calculation:

In $\Delta ABC$ , the sum of all interior angles is equal to 180 degrees.

  $\begin{array}{l}m\angle CAB+m\angle CBA+m\angle BCA=180°\\ m\angle CAB+90°+23°=180°\left[\because \overline{AB}\perp \overline{BC}\right]\\ m\angle CAB=180°-90°-23°\\ m\angle CAB=67°\end{array}$

In $\Delta ABD$ , the sum of all interior angles is equal to 180 degrees.

  $\begin{array}{l}m\angle BAD+m\angle ADB+m\angle ABD=180°\\ 67°+90°+m\angle ABD=180°\left[\because \overline{BD}\perp \overline{AC}\right]\\ m\angle ABD=180°-90°-67°\\ m\angle ABD=23°\end{array}$

Therefore, the measure of $\angle ABD$ is 23°.

(c)

To determine

To Explain: Why $m\angle ABD$ always equal to $m\angle C$ .

Answer to Problem 23WE

  $m\angle ABD=23$

Explanation of Solution

Given Information: In the figure shown below, $\overline{AB}\perp \overline{BC}\text{ and }\overline{BD}\perp \overline{AC}$ .

  

Calculation:

In $\Delta ABC$ , the sum of all interior angles is equal to 180°.

  $\begin{array}{l}m\angle CAB+m\angle CBA+m\angle BCA=180°\\ m\angle CAB+90°+m\angle BCA=180°\left[\because \overline{AB}\perp \overline{BC}\right]\\ m\angle CAB=180°-90°-m\angle BCA\\ m\angle CAB=90°-m\angle BCA\\ m\angle CAB=90°-m\angle C\end{array}$

In $\Delta ABD$ , the sum of all interior angles is equal to 180°.

  $\begin{array}{l}m\angle BAD+m\angle ADB+m\angle ABD=180°\\ m\angle CAB+m\angle ADB+m\angle ABD=180°\\ \left(90°-m\angle C\right)+90°+m\angle ABD=180°\left[\because \overline{BD}\perp \overline{AC}\right]\\ m\angle ABD=180°-90°-90°+m\angle C\\ m\angle ABD=m\angle C\end{array}$

Therefore, $m\angle ABD$ is always equal to the $m\angle C$ .