Chapter 5.5, Problem 29PSC

a.

To determine

To find: The value of $a$ in terms of $x\text{ and }y$ .

Answer to Problem 29PSC

  $a={180}^0+x-y$

Explanation of Solution

Given information:

Given $m\parallel n$ .

  

Calculation:

Given $m\parallel n$ .

  

From above figure,

  $\begin{array}{l}m\parallel n\\ \text{and }a=x+\left({180}^0-y\right)\\ \Rightarrow a={180}^0+x-y\end{array}$

b.

To determine

To find: What must be true for $y-x$ if $a>{90}^0$ .

Answer to Problem 29PSC

  $\left(y-x\right)<{90}^0$

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

Given $m\parallel n$ .

  

From above figure,

  $\begin{array}{l}m\parallel n\\ \text{and }a=x+\left({180}^0-y\right)\\ \Rightarrow a={180}^0+x-y\\ \Rightarrow y-x={180}^0-a\\ \text{when }a>{90}^0\\ \text{then }\left(y-x\right)<{90}^0\end{array}$