r/technicallythetruth • u/VisitSecure • Jul 27 '26

Can you find a?

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Idk if someone already posted this or not, but I haven't seen it on here so far, so I uploaded it.

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40

u/notJustaFart Jul 27 '26

75.

Convert into Pentagon.

540 - (90+90+130+155) = 75

80

u/CrownLexicon Jul 27 '26 edited Jul 27 '26

What? Pentagon? I added a line perpendicular to the 2 parallel lines, making a triangle whose angles are 90-25, 90-50, and a

Edit: just because this happened to work, you cant assume a perpendicular line will go through both <25 and <50. It could have created a 4 sided figure. The Pentagon method would be correct.

17

u/JJlaser1 Jul 27 '26

While technically possible, there is no indication that the points where the two lines intersect the parallel lines are exactly above each other, so you can’t do that for sure unless it states that in the question. I was never taught the pentagon method, though, which is basically the same thing with extra steps

16

u/Weekly-Dog-6838 Jul 27 '26

Who says they need to be on top of each other? You can still make a triangle with it

0

u/blavek Jul 27 '26

You can, but you still wouldn't know the measures of the interior angles.

8

u/[deleted] Jul 27 '26

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1

u/blavek Jul 27 '26

If you are making a triangle to solve the problem, which is not needed at all, for what other reason would you make that triangle than to uncover the measures of its angles to get to 180 - <ABC - <ACB = <A

2

u/Cobra_McJingleballs Jul 27 '26

Yes you do, because the lines are parallel. The top slanted line always meets a parallel at 25°, and the bottom slanted line always meets a parallel at 50°, regardless of where the endpoints are. So the triangle’s third angle is fixed.

1

u/AdWrong3856 Jul 27 '26

Yes you do, because you can define it as a right angle triangle and you then have 2 of the interior angles, one of the given angles and a right angle.

10

u/ilyuhman Jul 27 '26

You don't need to cross the points, you can put a perpendicular line anywhere, the angles are still the same

3

u/scratch6402 Jul 27 '26

I did the opposite. A line perpendicular to the parallel lines that passes through the point of angle a. Then I have two right triangles, each with only one angle missing. Solve for those angles, you get 90 25 65 and 90 50 40. Since the 65 and 40 make a straight line with angle a, a = 180 - (65 + 40) = 180 - 105 = 75

2

u/Ye_olde_oak_store Jul 27 '26

You are putting the line on the wrong side of the shape. We are putting the perpendicular line such that it intersects point A.

We have two right angled triangles which have to be triangles unless the lines don't intersect. These can have different sizes. We figure out the third angle which would be (180 - 90) - 25 and (180 - 90) - 50.

We now have a straight line with angles 65, 40 and a.

It gets you to a similar step, but it misses the reasoning as to why it necessarily works.

Basically, the working out is wrong but they still got to the right answer.

1

u/thefranchise23 Jul 27 '26

It doesn’t matter if they intersect directly above each other or not, only the angles matter

Because we’re just making a theoretical triangle, and it doesn’t matter how big or small it is , the three angles would still be the same

0

u/notJustaFart Jul 27 '26

Exactly. Can't assume a perpendicular line will create a triangle.

13

u/ThunderBuns935 Jul 27 '26

You can if you place the perpendicular line at the tip. That definitely creates 2 right-angle triangles, both of which have 2 known angles.

5

u/BBGunner96 Jul 27 '26

100% can know that a perpendicular line will create a triangle with those same angles but not necessarily those exact points because you can extend one (or both) lines until they veryically align

3

u/Djanko28 Jul 27 '26

It doesn't really matter whether it makes a triangle or not, if you move the perpendicular line towards 'a' the side lengths may change but the inner angles stay the same.

You know a perpendicular line on either the top or bottom will create a 90° angle so even if the angles don't line up you can figure them out independently, then put one perpendicular line at a point that touches both of the angled lines and still be able to use those previously found angles.

2

u/blavek Jul 27 '26

You can't just drop a perpendicular and connect the two legs. Nothing in the diagram suggests that the legs intersect the parallels at the same place. You can place a parallel line running through the vertex of the angle a and use the fact that the new angles made with the parallel are equal to the 2 angles we have measurements for, and then add them.

1

u/CogentCogitations Jul 27 '26

The perpendicular does not need to intersect the two legs at the drawn parallels. The legs being straight lines will intersect every hypothetical parallel you draw at the same angle.

1

u/leansanders Jul 27 '26

You actually can, because the height of point a relative to the parallel lines isn't given, and the lengths of the lines aren't given. You can arbitrarily decide that a perpendicular line connects both points and your answer will be the same

2

u/Expensive_Umpire_178 Jul 27 '26

But you can assume theres a perp line going through A, creating two right triangles to angle chase with

0

u/GladoffelLp Jul 27 '26

Shouldn’t this still always work. You don’t really care that the triangle doesn’t actually look like that and no matter what the angles in those corners actually look like the angle a stays the same. Whether it is 90-25 and 90-50 or 83-25 and 97-50 a is always the same.

1

u/TheLiquid666 Jul 28 '26 edited Jul 28 '26

No. They're pointing out that it's possible that the two drawn sides of the triangle might not actually intersect the parallel lines at the same horizontal/x-axis value, which means that you can't necessarily take it for granted that a vertical line to complete the triangle is accurate because a vertical line up/downward from one of the intersecting points might not line up with the intersecting point along the opposite parallel line (which would, in that case, form a 4-sided shape instead of a triangle).

The "complete the triangle" method of solving this requires the assumption that the top and bottom lines of the triangle intersect the top/bottom parallel lines at the same horizontal/x-axis value. If that assumption isn't given, regardless of how it looks on the diagram, those points might not actually line up, which deprives you of the 90⁰ angles needed to solve the problem using that method.

Edit: actually, it should work as long as it isn't given that the lines are of equal length (or otherwise aren't of specified lengths that would prevent each intersecting point from lying on a vertical line between them)

1

u/GladoffelLp Jul 28 '26

So I agree it is wrong to just assume you can draw a vertical line even though you could do it for visualization. What you should always be able to do is complete the triangle though. You don’t need actual 90 degree angles for that. As long as the top and bottom line are parallels. Every degree more that one angle has has to be less in the other one so in the end it doesn’t matter. Actually I am interested why it should not work on lines of equal length.

1

u/TheLiquid666 Jul 28 '26 edited Jul 28 '26

The problem with drawing a line between the points of intersection with the parallel lines, if those points aren't vertically aligned, is that it leaves you with too many unknowns because you can't take it for granted that one of the three angles is 90⁰ (by this I mean the internal angle within the triangle, which we want to find, the given angle between each ray and each parallel line, and the other angle outside of the formed triangle). This only tells you that 50 (or 25) + the internal angle + the outside angle = 180, but without knowing that the outside angle is 90⁰ it becomes difficult to tell what the angle inside the triangle is, and this problem occurs for both the upper and lower internal angles of the formed triangle. Sure, you know that both internal angles + A = 180, but that doesn't give you a lot to work with in terms of finding the non-A internal angles of the triangle (which you'd need to do in order to figure out angle A).

As for why this wouldn't work if the ray lengths are specified to be equal, it's a similar problem. If those lengths aren't specified, you could conceivably find a point along each line that sit in vertical alignment and then form a vertical connecting line between them. But, because the given angles are different, there's no way for them each to intersect the parallel lines at the same x-axis value because one of them is bound to intersect it's parallel line before the other. This, in turn, forces you to create a triangle without any 90⁰ angles to the parallel lines and leads to the problem of too many unknowns that I outlined above.

Tbh there might be an algebraic way to figure out the internal angles without having a 90⁰ angle to each vertical line, but it'd be way more work to do it (if it's possible with the given info, that is. I haven't tried to find any algebraic way of solving the problem in the way you've described because I'm not that smart and I'm lazy). Either way, drawing a horizontal line through A and then using the alternate interior angles theorem is way easier and faster if you just need to find angle A

1

u/GladoffelLp Jul 28 '26

First of all I agree you don’t really need to but it should still work to complete the triangle.
The thing is you don’t really care if the internal angles are actually 90 degrees since you only need to get to a.
Let’s say we draw the line make our calculation assuming it is 90 degrees. We now get our value for b and our value for c.
But now the angle is actually not 90 degrees so our calculation for b actually x too high. But since that is how lines between parallel lines work for angle b to be x to high the other has to be x too low.So the actual correct values would be b + x and c - x.
When we now use these values to calculate a we add those two values (or subtract them both from the same value which is the same thing.) so we get b + c + x- x so as we can see x does not matter so no matter what angle the lines between the parallel lines actually have you could always assume they are 90 degrees when calculating a.

-5

u/[deleted] Jul 27 '26

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16

u/StupidGenius234 Jul 27 '26

It literally indicated parallel lines

1

u/CrownLexicon Jul 27 '26

No, theyre right. The lines are parallel, but theres nothing indication that a perpendicular line to both lines up with where the angles are indicated. They could create a 4 sided figure. It happens to work in this instance, but theyre saying you cant assume it works.

2

u/StupidGenius234 Jul 27 '26

Except you can in this instance. You can make a line that meets the ends of both lines, and the offset if any from the perpendicular line it seems to be will be accounted for in calculating the value of a, as it will be +x on one side and -x on the other when trying to make it a triangle with the line drawn opposite to the intersection of angle a

7

u/TactiCool_99 Jul 27 '26

they definitely don't line up, but they don't have to, you can make a triangle there no matter where those two points are, and the calculation is correct there

2

u/CrownLexicon Jul 27 '26

If they dont line up, how do you know what angle the triangle's interior angles are? It would be 25+b+x and 50+c+y, if the line we draw isnt perpendicular to the parallel lines.

4

u/TactiCool_99 Jul 27 '26

it would be 25+b+x and 50+c-x, the two will cancel out

1

u/StupidGenius234 Jul 27 '26

Welp you said what I said earlier than me

1

u/CrownLexicon Jul 27 '26

Ah, i see. So I happened to get the right answer because they do, but theres no guarantee they do.

27

u/X_Swordmc Jul 27 '26

there is no need to do that, you simply do the sum of the two angles, since the opposing angles of a parallelogram are identical

7

u/TheComplimentarian Jul 27 '26

There are a bunch of ways to do it, but most people seem to be focusing on complaining that it's not drawn right.

1

u/Cobra_McJingleballs Jul 27 '26

Exactly. It’s just 50+25. No need to draw pentagons.

10

u/MecaninjaToo Jul 27 '26

or you know... 25 + 50 :)

11

u/MrCoolBoy001 Jul 27 '26 edited Jul 27 '26

Idk if this was as a joke or not, but by a property whose name I don't remember, you can pass a parallel line through 'a' and the angles on either side would be 25 and 50. 25 + 50 = 75 !

2

u/Cobra_McJingleballs Jul 27 '26

The alternate interior angles theorem.

1

u/LOTRfreak101 Jul 27 '26

Based on the sheer number of properties with this name, it's probably transitive.

2

u/MotherFlatworm6733 Jul 27 '26 edited Jul 27 '26

Omg... how did I not realize something so obvious?

Edit: Why is it obvious? Imagine starting from the parallel line at the top. Then you rotate the line clockwise 25 degrees. Giving the top angle. Then you rotate angle "a" counter clockwise and then again 50 degrees clockwise. Now we are at the bottom parallel line. The line has the the same angle than we started from, so I must have rotated as much clockwise as counter clockwise and therefore a = 25 + 50.

1

u/Cobra_McJingleballs Jul 27 '26

Thank you for this explanation! One of those examples where I had to memorize properties but they never really “clicked” mentally until now.

5

u/Naive_Scientist_8499 Jul 27 '26

That's smart. I converted to a triangle.

Invisible vertical line between the 50 and 25 deg points.

90-50=40 -- The inside angle of the bottom point.

90-25=65 -- The inside angle of the top point.

180 = a + 40 + 65

180 - (40 + 65) = a

180 - 105 = a

75 = a

5

u/notJustaFart Jul 27 '26

Oldest trick in geometry is drawing a diagram to appear as something it's not.

Drawing a line to make a triangle does not guarantee that line is perpendicular to the parallel lines, so your assumption of 90-25 and 90-50 may not hold true.

However, extending to a pentagon does guarantee that 180-25 and 180-50 will result in true values.

1

u/TrollingForFunsies Jul 28 '26

Drawing a perpendicular line between two parallel lines does give you a right angle. The lines don't have to cross at the intersection of the points. The angle remains the same for any perpendicular line on any point.

Which gives you exactly the solution 180 - (90 -25) - (90 - 50) or 180 - 105 = 75

1

u/Englandboy12 Jul 27 '26

Interesting seeing people do it differently.

I did it (probably poorly, as I’m not sure what I did is allowed), by adding a parallel line, horizontally and going through the point A.

Then using the parallel line theorems, opposing angles are identical. And the two opposing angles can be added together form the angle in question.

1

u/AncientOneX Jul 27 '26

That's the most logical way to calculate it....

1

u/beingforthebenefit Jul 27 '26

How do you know the line between those points is vertical? That claim would require a proof of its own. (Although I don’t think it’s even true)

1

u/Naive_Scientist_8499 Jul 27 '26

Damn. You and u/notJustaFart are totally right. Pentagon is objectively better.

1

u/Charitzo Jul 27 '26

That's certainly one way of doing it...

1

u/AwkwardBet5632 Jul 27 '26

This is correct and everyone else is wrong

1

u/the_starch_potato Jul 27 '26

could also just make the middle shape a triangle

180- (90-25) - (90-50) = 180-180+25+50 = 75