Video: AQA GCSE Mathematics Higher Tier Pack 4 β€’ Paper 3 β€’ Question 10

𝑂𝑃, 𝑄𝑅, and 𝑆𝑇 are three straight lines. a) If you assume that 𝑂𝑃 and 𝑄𝑅 are parallel, what is the size of angle 𝑦? b) 𝑂𝑃 and 𝑄𝑅 are not in fact parallel and the size of angle 𝑧 is 150Β°. How does this affect the size of angle 𝑦? Tick a box. 𝑦 is smaller. 𝑦 is the same. 𝑦 is larger. Show your working to justify your answer.

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Video Transcript

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

Let’s begin with part a) and we will come back to part b). So for part a), we are assuming that 𝑂𝑃 and 𝑄𝑅 are parallel, which sometimes can be denoted with arrows. And we are asked to find the size of angle 𝑦.

Let’s go ahead and say that 𝑂𝑃 and 𝑇𝑆 β€” these lines β€” intersect at point 𝐴 and 𝑄𝑅 and 𝑇𝑆 intersect at point 𝐡. We needed to add these points that way we could talk more specifically about the angles around the points. So if indeed 𝑂𝑃 and 𝑄𝑅 are parallel, then angle 𝑇𝐴𝑂 will be equal to angle 𝑇𝐡𝑄, so the two blue angles. And this is because these two angles are corresponding. And corresponding angles are congruent.

So we can set angle 𝑇𝐴𝑂 equal to angle 𝑇𝐡𝑄 because we have expressions for each of them. Angle 𝑇𝐴𝑂 is equal to six π‘₯ plus 10 and angle 𝑇𝐡𝑄 is equal to nine π‘₯ minus five. So now, let’s 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 𝑦. 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 we’re just talking about, they are angle 𝑇𝐡𝑄 and angle 𝑅𝐡𝑆. And these angles are vertically opposite. And vertically opposite angles are congruent. So we can set these angles equal to each other.

Angle 𝑇𝐡𝑄 is equal to nine π‘₯ minus five degrees. So we can remember to add the degrees at the end to the 𝑦 that we’re solving for and then set it equal to angle 𝑅𝐡𝑆 which is equal to 𝑦. 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 𝑦 is equal to 40. And like we said, we need to add degrees. So if we assumed that 𝑂𝑃 and 𝑄𝑅 were parallel, the size of angle 𝑦 would be 40 degrees. So to make room for part b), let’s go ahead and erase this. But answer the question for part a). So we need to write 40 degrees for part a). And again, we’re doing this to create room for part b). So the answer to part a) was the size of angle 𝑦 is 40 degrees.

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

So a couple of things are different for part b). 𝑂𝑃 and 𝑄𝑅 are not parallel. So now, we actually don’t know that these lines are parallel. But we do know the size of angle 𝑧; it’s equal to 150 degrees. Now since 𝑂𝑃 and 𝑄𝑅 aren’t parallel, we can’t use corresponding angles and some of those properties that come from parallel lines.

But in the original statement, we were told that 𝑂𝑃, 𝑄𝑅, and 𝑇𝑆 or 𝑆𝑇 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 𝑧 is equal to 150 degrees, angle 𝑂𝐴𝑇 plus angle 𝑇𝐴𝑃 should equal 180 degrees because like we said the angles on a straight line sum to 180 degrees. Angle 𝑂𝐴𝑇 is equal to six π‘₯ plus 10 degrees and angle 𝑇𝐴𝑃 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 𝑦.

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 don’t need parallel lines for these to be vertically opposite. Notice that the two green angles don’t even touch line 𝑂𝑃. So as we used in part a), angle 𝑇𝐡𝑄 will be equal to angle 𝑅𝐡𝑆 because they’re vertically opposite.

So we can say that 𝑦 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 𝑦? Well, now we have found that the size of angle 𝑦 is actually 25 degrees. Originally, we said that the size of angle 𝑦 was 40 degrees. And now, we say that it’s only 25 degrees. So we have found that the size of angle 𝑦 is smaller. So we will tick the box that says 𝑦 is smaller.

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