Chapter 15.2, Problem 20PSC

a.

To determine

To list: the given sides $\overline{AC},\overline{CB},\overline{AB},\overline{AD}$ in increasing order.

Answer to Problem 20PSC

  $\overline{AD}<\overline{CD}<\overline{AC}<\overline{BC}<\overline{AB}$ .

Explanation of Solution

Given:

The angles $\angle ACB\text{ and}\angle CDB$ are right angles with $\angle B={20}^{\circ }$ .

  

Calculation:

According to given figure,

  $\begin{array}{l}\angle A={180}^{\circ }-\left({20}^{\circ }+{90}^{\circ }\right)\\ \angle A={180}^{\circ }-{110}^{\circ }={70}^{\circ }\end{array}$

  $\begin{array}{l}\angle DCA={180}^{\circ }-\left({90}^{\circ }+{70}^{\circ }\right)\\ \angle DCA={180}^{\circ }-{160}^{\circ }={20}^{\circ }\end{array}$

Now, in $\Delta DCA$ , $\angle DCA$ is the smallest angle, so $\overline{AD}$ is the smallest segment.

Similarly, in $\Delta DCB$ , $\angle B$ is the smallest angle, so $\overline{CD}$ is the smallest segment.

Now, comparing $\overline{CD}$ to $\overline{AD}$ in $\Delta DAC$ , $\angle A$ is the opposite of $\overline{DA}$ is the smallest segment.

The other segment can be put in order in this same manner:

  $\overline{AD}<\overline{CD}<\overline{AC}<\overline{BC}<\overline{AB}$

Hence, the list in increasing order is $\overline{AD}<\overline{CD}<\overline{AC}<\overline{BC}<\overline{AB}$

b.

To determine

To adjust: the location of $\overline{DB}$ in the list.

Answer to Problem 20PSC

  $\overline{AC}<\overline{DB}<\overline{BC}$ .

Explanation of Solution

Given:

The angles $\angle ACB\text{ and}\angle CDB$ are right angles with $\angle B={20}^{\circ }$ .

  

Calculation:

Using trigonometry,

In $\Delta BDC$ , $\angle BDC={70}^{\circ }\Rightarrow \sin {70}^{\circ }=\frac{BD}{BC}$ .

Again, in $\Delta ABC$ , $\angle ABC={70}^{\circ }\Rightarrow \tan {20}^{\circ }=\frac{AC}{BC}$ .

  $\begin{array}{l}\text{since, }\sin {70}^{\circ }>\tan {20}^{\circ }\\ \text{therefore},\frac{BD}{BC}>\frac{AC}{BC}\\ \Rightarrow BD>AC\end{array}$

Thus, $\overline{DB}$ would fit either right before or right after $\overline{AC}$ . Then the order are $\overline{AC}<\overline{DB}<\overline{BC}$