Question

what must be the value of x so that lines a and b are parallel lines cut by transversal f? 10 20 22 32 (6x - 36)° 96°

Explanation:

Step1: Use corresponding - angles property

When two parallel lines are cut by a transversal, corresponding angles are equal. Here, the angle of measure $96^{\circ}$ and the angle of measure $(6x - 36)^{\circ}$ are corresponding angles. So, we set up the equation $6x-36 = 96$.

Step2: Solve the equation for $x$

Add 36 to both sides of the equation: $6x-36 + 36=96 + 36$, which simplifies to $6x=132$.
Then divide both sides by 6: $\frac{6x}{6}=\frac{132}{6}$, so $x = 22$.

Answer:

C. 22