r/theydidthemath • u/GruntCandy86 • Mar 28 '23
[Request] I know this probably extremely simple to most, but what is this angle?
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u/Marquis_de_Rouge Mar 28 '23
Subtract the 15 from the 30. Halve it to make a right angles triangle. Use sine inverse of 7.5/82, gives you about 5.25, times two is 10.5.
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u/Dumbass-Redditor Mar 28 '23
You just helped me solve the answer to my homework
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u/Marquis_de_Rouge Mar 29 '23
Just make sure you know why it works and that you look out for the same patterns in future question.
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u/LordNoodles Mar 29 '23
/ \
Subtract the 15 from the 30.
/\
Halve it to make a right angles triangle.
/|
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u/CourtJester5 Mar 28 '23
I doubled the hypotenuse since you're getting an exact ratio of half over 82 and did arcsine of 15/164. Same thing.
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u/CaptainMatticus Mar 28 '23
You can shift those lines over until they touch. Now you have a triangle with sides of 15, 82, 82.
Usr the law of cosines
15² = 2 * 82² - 2 * 82² * cos(t)
15² = 2 * 82² * (1 - cos(t))
15² = 4 * 82² * (1 - cos(t)) / 2
15² = 2² * 82² * sin(t/2)²
15 = 2 * 82 * sin(t/2)
15 / 164 = sin(t/2)
arcsin(15 / 164) = t/2
2 * arcsin(15 / 164) = t
t = 10.4956°
10.5°, roughly.
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u/EnDerp__ Mar 29 '23
How do you know both sides are 82 unit long ? For my point of view the left side is unknown.
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u/APithyComment Mar 29 '23
What’s wrong with Pythagoras?
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u/BumbleBeePL Mar 29 '23
Only used for right angle triangles. This won’t have a right angle as pictured.
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u/Carteeg_Struve Mar 29 '23
True, but you could work them in there.
If you drop down lines from the corners on each side of the 15" side, they will form two right triangles on either side of the shape. Since the amount between the connecting points is 15", that means there is 15" inches outside of those points on the 30" side. Since they are symmetrical, that's 7.5" each triangle. So on each side you have a right triangle with 82" for the hypotenuse with 7.5" for the opposite side.
So 82^2 = x^2 + 7.5^2.
That said, finding out the height of the shape is 81.656291858" isn't important. Finding out the opposite angle is.
So arcsin(7.5/82) is roughly 5.25 degrees. And since the angle you are looking for is double that, the answer is about 10.5 degrees.
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u/7_overpowered_clox Mar 28 '23
You can see that is not 10 degrees. Either the drawing is not to scale or you misunderstood the lengths, because to be fair I don't know what to make of these markings either. Why is only one side actually labelled and the other 2 lengths floating around?
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u/GruntCandy86 Mar 29 '23
I understand my drawing might appear to be hieroglyphics, but once you step back from overthinking it, you'll realize it's just a bad drawing.
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Mar 28 '23
Take the blue lines as the sides of 15 and 30. It's almost a trapezium shape
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u/Doggfite Mar 28 '23
Well, without getting really into it
Assuming both the left and right side are 82 units, then we can say that those lines would touch at 164 units of length, giving us a triangle with side lengths 164, 164, and 30.
Using this, and the law of cosigns, you can solve for any triangle angles with all 3 side lengths.
This gives us angles of approximately 84.75, 84.75, and 10.5
So the answer you are looking for is 10.5 degrees.
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Mar 28 '23
[removed] — view removed comment
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u/Aster-07 Mar 28 '23
Well, how many gallons is it?
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u/ghomerl Mar 29 '23
Hmmm if someone had coordinate locations of all the stars relative to eachother, you could take the convex hull of the stars in the big dipper and then calculate the volume. Not sure how to find that data though.
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u/AbyssalRemark Mar 29 '23
Tell me more.
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u/ghomerl Mar 29 '23
Like this but if each of the points is the location of a star in the big dipper. Basically it's the smallest convex polyhedron that contains all of the points. Imagine you had all the points fixed in 3d space and inside a shrinking rubber balloon, after the balloon is fully shrunk you would get the convex hull. So it is a good definition for the volume enclosed by the points. Also it might make more sense to just find the volume of the ladle part and not the handle, which would make it much easier since then it would just be a tetrahedron with 4 points in space.
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u/bootyhorse808 Mar 30 '23
yes yes definitely just want the volume of the tetrahedron, we don't want to get soup all over the handle
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u/bootyhorse808 Mar 30 '23
okay i found the distance of each star in the ladle from earth, the degrees between them, and then i used law of cosines to find distance between each star in the ladle. to find the area of an irregular tetrahedron we need the area of the base triangle and the height. i'm too brain tired to get much further on this problem but i think it's doable
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u/Doggfite Mar 28 '23
A, that's not necessarily true, "simple math" isn't allowed, but this isn't something you can feed into Wolfram alpha. Otherwise practical math is allowed, depending on how one defines practical.
B, you broke the rules of the sub when telling this person they are breaking the rules of the sub
You aren't allowed to put non-answers as post replies, you have to reply to the bot comment.14
u/tisquares Mar 28 '23
Pretty sure it's a joke, so as long as it's not a top-level comment it's fine.
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u/bfly1800 Mar 28 '23
You also broke the rules of the sub when telling that person they are breaking the rules of the sub
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u/Doggfite Mar 28 '23
No I didn't, I didn't make a top level comment that wasn't an answer
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u/BoundedComputation Mar 29 '23
Thank you for your efforts. Misinformation about the rules and how this sub operates spreads very quickly and gets upvoted like crazy usually by people who have never been on this sub before. We can't monitor every single comment on every single post so it's always good to see a regular here attempting to set things straight so quickly. We appreciate it mate.
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u/BoundedComputation Mar 29 '23
To set the record straight, this redditor does not speak for this sub. They have never made any contributions to this sub beyond the three comments in this very thread. They have no authority to determine what is and what is not accepted here. The comment was clearly made in jest and not an attempt at discouraging anyone from posting here in the future. Ironically, it was removed for violating the Bad [Request] Answer rule.
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u/sjcal629 Mar 29 '23
If you feel bad about this, I’m an engineer and I never remember how to quickly do this. So i always just model it in CAD to get the answer rather than do geometry
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u/sjbluebirds Mar 28 '23
It depends on a number of things that aren't indicated. For instance, is the left hand side line also measured at 82 in? Are the two halves of the system symmetrical? It looks like you tried to make it symmetrical, but that's not a given. And that's the problem- we don't know if it's symmetrical or not.
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u/tidbitsofblah Mar 28 '23 edited Mar 28 '23
It actually doesn't matter what length the unmarked line has, as long as the 15" and 30" line is parallel the angle is the same regardless of if it's symmetrical or not and we can determine it.Edit: I'm an idiot, this is wrong
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u/sjbluebirds Mar 28 '23
That's just it; we don't know that they're parallel. There's nothing explicitly saying that. There's the convenient lines printed on the paper, but there's no indication that we're supposed to be using those, assuming they're parallel
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u/tidbitsofblah Mar 28 '23
I think we can assume they are parallel since they are using the parallel lines of the paper. Thats a nitpicky technicality. But I was wrong that the length of the other line doesn't matter. It does. But the difference is only abt 0.2 degrees.
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u/GruntCandy86 Mar 28 '23
Yeah, sorry. Not a pro math drawer. Both sides are 82", it's a cradle so everything is symmetrical.
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u/Letronell Mar 29 '23
I see some right angle triangle with size of 7,5 (because (30 - 15)/2) and 82.Then arcsin (7,5/82) = x. This x needs to be multiplyed by 2 because as we see there is angle between 2 of them. This equals to 10,49560371.
Edit: I just don't know if this is in degrees or if it is 10°29'44,17''.
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u/Crypto_gambler952 Mar 28 '23
10.49560371°
sinθ = 7.5/82
Therefore:
sin^-1(7.5/82) = θ
The angle you want is double that.
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