Chapter 12.1, Problem 34E

a.

To determine

To measure each angle and length of each side in the $\Delta ABC$ and $\Delta DEC$ .

Answer to Problem 34E

The angle of each side is ${90}^{\circ }$ , ${60}^{\circ }$ and ${30}^{\circ }$ .

Explanation of Solution

Given:

In $\Delta ABC$ and $\Delta DEC$ external angle $\angle ACE=\angle BCD={150}^{\circ }$ . And the length $BA=4$ , $DE=2$

Concept used:

Two angles are complementary angles if the sum of their measure in ${90}^{\circ }$ two angles are supplementary angles if their sum measure is ${180}^{\circ }$ .

Calculation:

From the diagram

  $\begin{array}{l}\Rightarrow \angle ACB+\angle DCE={180}^{\circ }\\ \Rightarrow {150}^{\circ }+\angle DCE={180}^{\circ }\\ \Rightarrow \angle DCE={180}^{\circ }-{150}^{\circ }\\ \Rightarrow \angle DCE={30}^{\circ }\end{array}$

  $\Rightarrow \angle C={30}^{\circ }$

So, in $\Delta ABC$

  $\Rightarrow \angle CAB+\angle ABC+\angle BCA={180}^{\circ }$

  $\Rightarrow \angle A+\angle B+\angle C={180}^{\circ }$

  $\begin{array}{l}\Rightarrow \angle A+{90}^{\circ }+{30}^{\circ }={180}^{\circ }\\ \Rightarrow \angle A={180}^{\circ }-{120}^{\circ }\\ \Rightarrow \angle A={60}^{\circ }\end{array}$

And similarly, in $\Delta DEC$ .

  $\Rightarrow \angle DCE+\angle CED+\angle EDC={180}^{\circ }$

  $\begin{array}{l}\Rightarrow \angle C+\angle E+\angle D={180}^{\circ }\\ \Rightarrow {30}^{\circ }+{90}^{\circ }+\angle D={180}^{\circ }\\ \Rightarrow \angle D={180}^{\circ }-{120}^{\circ }\\ \Rightarrow \angle D={60}^{\circ }\end{array}$

Thus, the angles are $\angle A={60}^{\circ }$ , $\angle C={30}^{\circ }$ , $\angle D={60}^{\circ }$ .

Since,

  $\angle A={60}^{\circ }$ , $\angle D={60}^{\circ }$ and $\angle E={90}^{\circ }$ , $\angle B={90}^{\circ }$

Therefore,

  $\Delta ABC$ and $\Delta DEC$ are similar, by angle angle theorem.

  $\begin{array}{l}\Rightarrow \frac{AB}{DE}=\frac{BC}{CE}\\ \Rightarrow \frac{4}{2}=\frac{BC}{CE}\\ \Rightarrow \frac{2}{1}=\frac{BC}{CE}\\ \Rightarrow BC=2\text{ and }CE=1\end{array}$

Since, it is aright angled triangle,

Therefore,

  $AC=\sqrt{A{B}^2+B{C}^2}$

  $\begin{array}{l}\Rightarrow AC=\sqrt{4^2+2^2}\\ \Rightarrow AC=2\sqrt{5}\end{array}$

Also,

  $CD=\sqrt{C{E}^2+E{D}^2}$

  $\begin{array}{l}\Rightarrow CD=\sqrt{1^2+2^2}\\ \Rightarrow CD=\sqrt{5}\end{array}$

Conclusion:

The angle of each side are $\angle A={60}^{\circ }$ , $\angle C={30}^{\circ }$ , $\angle D={60}^{\circ }$ and the length of each side are $BC=2\text{ and }CE=1$ , $\begin{array}{l}AC=2\sqrt{5}\\ \end{array}$ and $CD=\sqrt{5}$

b.

To determine

To compare:How $\Delta ABC$ and $\Delta DEC$ are related.

Answer to Problem 34E

  $\Delta ABC$ and $\Delta DEC$ are related to each other by angle angle theorem.

Explanation of Solution

Concept Used:

When two angles of a triangle are same then by angle angle theorem they are called as similar triangles.

Since in $\Delta ABC$ and $\Delta DEC$

  $\angle A={60}^{\circ }$ , $\angle D={60}^{\circ }$ and $\angle E={90}^{\circ }$ , $\angle B={90}^{\circ }$ .

Therefore, by angle angle theorem $\Delta ABC$ and $\Delta DEC$ are similar.

Conclusion:

  $\Delta ABC$ and $\Delta DEC$ are related to each other.