Chapter 14.4, Problem 36WE

a.

To determine

To find: What rotation maps $A$ to $B$ & $D$ to $E$ .

Answer to Problem 36WE

It is rotation $R_{C,-60}$ for ${60}^{°}$ clockwise about point $C$ .

Explanation of Solution

Given:

  $\Delta ABC$ and $\Delta DCE$ are equilateral.

Concept Used:

A rotation is a transformation suggested by rotating a paddle wheel. When wheel stops, new position of a paddle referred as the image of initial position.

Calculation:

As per the given problem

  

  $\Delta ABC$ and $\Delta DCE$ are equilateral.

It is rotation $R_{C,-60}$ for ${60}^{°}$ clockwise about point $C$ .

  

Hence,

It is rotation $R_{C,-60}$ for ${60}^{°}$ clockwise about point $C$ .

b.

To determine

To explain: $AD\widetilde{-}BE$

Answer to Problem 36WE

  $AD\widetilde{-}BE$

Explanation of Solution

Given:

  $\Delta ABC$ and $\Delta DCE$ are equilateral.

Concept Used:

A rotation is a transformation suggested by rotating a paddle wheel. When wheel stops, new position of a paddle referred as the image of initial position.

Calculation:

As per the given problem

  

  $\Delta ABC$ and $\Delta DCE$ are equilateral.

It is rotation $R_{C,-60}$ for ${60}^{°}$ clockwise about point $C$ , maps $A$ to $B$ & $D$ to $E$ .

  

Since rotation is an isometry, we can conclude $AD\widetilde{-}BE$

Hence,

  $AD\widetilde{-}BE$

c.

To determine

To find: Measure acute angle between $\overleftrightarrow{AD}$ & $\overleftrightarrow{BE}$

Answer to Problem 36WE

Acute angle between $\overleftrightarrow{AD}$ & $\overleftrightarrow{BE}$ is ${60}^{°}$ .

Explanation of Solution

Given:

  $\Delta ABC$ and $\Delta DCE$ are equilateral.

Concept Used:

A rotation is a transformation suggested by rotating a paddle wheel. When wheel stops, new position of a paddle referred as the image of initial position.

Calculation:

As per the given problem

  

  $\Delta ABC$ and $\Delta DCE$ are equilateral.

It is rotation $R_{C,-60}$ for ${60}^{°}$ clockwise about point $C$ , maps $A$ to $B$ & $D$ to $E$ .

Since $\overleftrightarrow{AD}$ is rotated for ${60}^{°}$ , acute angle between $\overleftrightarrow{AD}$ & $\overleftrightarrow{BE}$ is ${60}^{°}$ .

  

Hence,

Acute angle between $\overleftrightarrow{AD}$ & $\overleftrightarrow{BE}$ is ${60}^{°}$ .