### Video Transcript

ππ, ππ
, and ππ are three straight lines. Part a) If you assume that ππ and ππ
are parallel, what is the size of angle π¦? Part b) ππ and ππ
are not in fact parallel and the size of angle π§ is 150 degrees. How does this affect the size of angle π¦? Tick a box. π¦ is smaller, π¦ is the same, π¦ is larger. Show your working to justify your answer.

Letβs begin with part a) and we will come back to part b). So for part a), we are assuming that ππ and ππ
are parallel, which sometimes can be denoted with arrows. And we are asked to find the size of angle π¦.

Letβs go ahead and say that ππ and ππ β these lines β intersect at point π΄ and ππ
and ππ intersect at point π΅. We needed to add these points that way we could talk more specifically about the angles around the points. So if indeed ππ and ππ
are parallel, then angle ππ΄π will be equal to angle ππ΅π, so the two blue angles. And this is because these two angles are corresponding. And corresponding angles are congruent.

So we can set angle ππ΄π equal to angle ππ΅π because we have expressions for each of them. Angle ππ΄π is equal to six π₯ plus 10 and angle ππ΅π is equal to nine π₯ minus five. So now, letβs solve. If we subtract six π₯ from both sides of the equation, we get 10 equals three π₯ minus five. And then, if we add five to both sides of the equation, we have 15 is equal to three π₯. And finally, dividing both sides by three, we get that π₯ is equal to five. And since π₯ is a part of the angle, we can say that π₯ is equal to five degrees.

Now, our goal is to find the size of angle π¦. So we need to find an angle that is somehow related to this one. Well, the nine π₯ minus five degrees angle is directly across from it or diagonal. And these two angles that we weβre just talking about, they are angle ππ΅π and angle π
π΅π. And these angles are vertically opposite. And vertically opposite angles are congruent. So we can set these angles equal to each other.

Angle ππ΅π is equal to nine π₯ minus five degrees. So we can remember to add the degrees at the end to the π¦ that weβre solving for and then set it equal to angle π
π΅π which is equal to π¦. So in order to solve here, we have to know what to plug in for π₯. And we know what to plug in for π₯, five.

So we take nine times five and then subtract five. Nine times five is 45 and 45 minus five is 40. So π¦ is equal to 40. And like we said, we need to add degrees. So if we assumed that ππ and ππ
were parallel, the size of angle π¦ would be 40 degrees. So to make room for part b), letβs go ahead and erase this. But answer the question for part a). So we need to write 40 degrees for part a). And again, weβre doing this to create room for part b). So the answer to part a) was the size of angle π¦ is 40 degrees.

Part b) ππ and ππ
are not in fact parallel and the size of angle π§ is 150 degrees. How does this affect the size of angle π¦? Tick a box. π¦ is smaller, π¦ is the same, π¦ is larger. Show your working to justify your answer.

So a couple of things are different for part b). ππ and ππ
are not parallel. So now, we actually donβt know that these lines are parallel. But we do know the size of angle π§; itβs equal to 150 degrees. Now since ππ and ππ
arenβt parallel, we canβt use corresponding angles and some of those properties that come from parallel lines.

But in the original statement, we were told that ππ, ππ
, and ππ or ππ are all three straight lines. And angles on a straight line sum to 180 degrees. So we could use that. So if we know that π§ is equal to 150 degrees, angle ππ΄π plus angle ππ΄π should equal 180 degrees because like we said the angles on a straight line sum to 180 degrees. Angle ππ΄π is equal to six π₯ plus 10 degrees and angle ππ΄π is equal to 150 degrees. And we can use this to solve for π₯. And once we find π₯, we can use it to help us find π¦.

10 plus 150 is equal to 160. And if I subtract 160 from both sides of the equation, we have that six π₯ is equal to 20. And then dividing both sides of the equation by six, we have that π₯ is equal to twenty sixths. However, we need to reduce this fraction. And 20 and six can both be reduced by two, making it ten thirds.

Now, luckily, the angles in green are still considered vertically opposite angles. We donβt need parallel lines for these to be vertically opposite. Notice that the two green angles donβt even touch line ππ. So as we used in part a), angle ππ΅π will be equal to angle π
π΅π because theyβre vertically opposite.

So we can say that π¦ is equal to nine π₯ minus five because vertically opposite angles are equal or they are congruent, except this time we will substitute ten thirds in for π₯. So we need to take nine times ten thirds. Three goes into itself once and into nine three times. So three times 10 over one or three times 10 is 30 and 30 minus five is 25. And like we said before, we need to add degrees.

So how does this affect the size of angle π¦? Well, now we have found that the size of angle π¦ is actually 25 degrees. Originally, we said that the size of angle π¦ was 40 degrees. And now, we say that itβs only 25 degrees. So we have found that the size of angle π¦ is smaller. So we will tick the box that says π¦ is smaller.