Question

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

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: $x=\frac{132}{6}=22$.

Answer:

22