Chapter 2.8, Problem 7PSA

To determine

Two given angles are possible

Answer to Problem 7PSA

Angle $\angle 1\left(=\left(3x-3\right)\right)°\text{ \& }\angle 2\left(=\left(4\text{x}\right)\right)°\text{ are not possible}$

Explanation of Solution

Given information :

Two lines are intersecting at one point and two angles are formed opposite to each other

Formula used:

Vertically opposite angles are always equal

Proof:

Since two lines are intersecting at one point

Then, angles formed opposite to each other will be equal

(because they are vertically opposite angles)

  $\begin{array}{l}\text{Let }\angle 1=\left(3\text{x}-3\right)\text{and }\angle \text{2=}\left(4\text{x}\right)°\\ \Rightarrow \angle 1=\angle 2\end{array}$

  $\begin{array}{l}\Rightarrow \left(3\text{x}-3\right)°\text{=}\left(4\text{x}\right)°\\ \Rightarrow 4\text{x}-3\text{x= }-\text{3}\\ \Rightarrow \text{x=}-3\end{array}$

It gives,

  $\begin{array}{l}\angle 1=\left(3\text{x}-3\right)°=-12°\text{ and}\\ \angle \text{2=}\left(4\text{x}\right)°=-12°\end{array}$

Hence, angles with (-ive) magnitude are not possible.