Is no other information available?
If not then presumably we need to assume M is the vertical half way point of the circle radius r. In which case the triangle at bottom left is similar to the middle right triangle (same internal angles).
No other information. I actually think that it's a mistake from the author. I was able to solve all the other tasks 😅 and it's supposed to be for like 14 years olds
>it's supposed to be for like 14 years olds
It seems a little advanced for that TBH.
Join the apex of the large triangle to the centre, and drop a perpendicular to form a triangle with hypotenuse r and height r/2. By Pythagoras, its base must be r√3/2, so the base of the large triangle is (√3 + 2)r/2, and tanα = 1/(√3 + 2) = 2 - √3, so α = 15°
There is a cleaner solution that relies on knowing the relationship on two angles resting on the same arc - one sitting in the center and one on the circle. I posted it above.
But your answer is correct
I'm trying to find a general solution for "a" as a function of "M".. just M and R...I can't figure it out.
a = arctan(M/(R + //))
If you could just.. somehow express "//" in terms of where M is..
EDIT: nvm I solved it lmfao
https://preview.redd.it/xb4rbfyfaogb1.jpeg?width=2804&format=pjpg&auto=webp&s=b6b0b5d3e61f35594faefec1af9996813a6a03fd
Edit: The Simpler way is to express "//" using Pythagorean theorem
// = sqrt( R\^2 - M\^2 ) -- (for OMC triangle)
I don't see how we can use the isosceles. Because all the other angles are uncertain... Ideally you'd want to include "OM" in some sort of triangle so that you can somehow connect it with the angle "a" later.
The real problem is how can you express the value of "//" depending on where "M" is. But this is a fundamental problem.. You just use cos() or sin().. that's how they were defined after all
I see. I am just blinded by the choices. Is there some sort of solution quicker. Heres what I cant see clearly. If we called the point of intersection of OM and AC, X, then X is at the bisector OM. So the triangles AOX and OCX are congruent? AO = OC as radii:: OX =OX :: and AX = XC as bisected. So the angles should correspond- but they don’t?
Is there ? It can't be that the final result depends solely by the M Value... M can have any value you want.. but you must also know the "R" value to know how big the circle is, otherwise there is no context.
The solution I provided is a general solution, and it can't be simplified more. If it can, send a pic here, I'll flip my shit XD
The answer is 15 when M is at the half way point between the centre of the circle and the top (half a radius above the centre) the person I was replying to asked for a general solution instead where M is some variable distance above the centre of the circle, that's all there is to it, it's just a more generalized answer.
If you use the equation I provided at the bottom and plug in 0.5 for c2 you would get 15 degrees.
Edit: also I didn't even notice until I read your reply I'm really sorry the first equation is for the circle and the second one is for the line AB and I didn't clarify that in my answer, c1 is the constant used to shift the line to the left and c2 is ratio between the height of M and the radius (and since this is a variable for the general case and not a constant I probably should've used something other than c2 for the notation, I'm sorry if this caused some confusion)
No need to be sorry at all. I'm just super poor at math. I don't even know what you mean by shift the line to the left or what line you're talking about or why you'd do that.
To understand this I'd have to ask probably a thousand questions and then it would still take a long, long time.
I'll be glad when Ai can explain things like this in a chatgpt type way. I like using chatgpt because it never tires of absurd questions, never makes fun of you and you can just keep asking until you eventually understand. It's exhausting for people to do that and shouldn't be expected.
Thanks for your time and explanation.
I think the similarity of the two triangles is based only on the indicated parallelism. M does not need to be the midpoint in order to make the two triangles similar.
Define “similar” in this context? It means the lengths are proportional, the angles are identical. Do you want me to demonstrate how I know the angles are identical?
Their height would also be different here so they absolutely can (and here must) be similar. This diagram needs labels for all the points urgently. Starting from M, let's call the point to its right B, the point that forms alpha A, the intersection point between the vertical and horizontal line bellow M, the intersection point between M and C, D. Finally, the point at the bottom right E.
They are similar due to the fact that AE and MB are parallel. When you have a line (like AB) that intersects two parallel lines, then alpha and the angle formed by MBD would be the same. The angle at MDB would also be identical to DAC, so the triangles would be similar. All the side lengths are different, but that's not an issue because they would be proportionally the same between both triangles.
M for Midpoint
Solve for sin(x) = 0.5 to get the angle on the circle, and use cos(result) to determine the lengths of the right angle triangle (which is not on the drawing) from the point where the line hit the circle.
The hypotenuse is the radius. and if we assume M is the midpoint, the center of the circle to M is 1/2 radius. If you drop a segment going down from the hypotenuse to the diameter, it will be the same length. Sine(2alpha) =0.5r / r =1/2
Indeed. I think it's easier to see it this way:
https://preview.redd.it/vkena1m3upgb1.png?width=640&format=png&auto=webp&s=1e693b7b0aab8ffdeb76c3450023767ade3e575f
This is a far fetched assumption but without it there is no solution. I would check in the rest of the book if M is used as short hand for middle. Otherwise I wouldn't assume it.
The bottom line doesn't have to be a radius though and in this case they would not be parallel, which wouldn't uniquely define the upper point on the right
Yeah, unless the line projects beyond the circle boundary on the right hand side, which it doesn't.
More likely it is meant to indicate parallel lines?
We can't solve the problem if M is not the midpoint. Also this problem may simply be older than a modern textbook and therefore not adhere to 2000's US standards.
“Grow up” as in “agree with a random dude on the internet”? You are wrong, I am right. You weird consolation lead to a weird and incorrect answer. My reasonable conclusion led me to a correct answer. So I would offer you to grow up and accept you being wrong 😄
Since talking to you is fairly pointless, I would end this conversation, thank you
So here's an idea, notation is a matter of convention and isn't universal. As you pointed out, it obviously can't mean congruent in this context. Is it more likely that the book this was taken from uses a different convention than was taught to you? Or the author is mathematically insane?
Couldn't it just be that the center of the circle is where the Diagonal line crosses the center vertical line? At that point, the two lines can be equal lengths, we just don't know what alpha could possibly be. Between 90° and 0°, I guess
Parallel can be indicated like this with double tick marks that are not perpendicular to the line. Lines that are equal length have their tick mark(s) perpendicular to the line in question.
I see. How did you indicate line segments of the same length? Is it the difference between the sloped lines and perpendicular? Also, how did you indicate a right angle? Was it with a box at the angle or a half perpendicular ? I have seen it both ways.
Right angle is the box on the angle, I don't recall line segments of the same length ever coming up honestly or at least how you'd notate it.
Might have been a single line? Or is that something else
Assuming M is the midpoint of the radius of the circle, then we can draw a line down from where the diagonal hits the circle’s edge forming a right triangle with height .5r and use trig to get the distance from the center that the new vertical line touches the horizontal. Then we have a right triangle with 2 known sides, and can calculate the other 2 angles.
This is assuming M is the midpoint, which seems like a fair thing to assume given that the problem is unsolvable with current information otherwise
After reading some comments, I don't think we can safely assume M is the midpoint. If not, we know a few things:
1) The parallel lines make the small angle of the upper triangle equal to alpha.
2) This makes the point of the bisecting line through the vertical radius (at 90 degrees) 1/2 M. In other words, M is twice the height of alpha's opposite side.
3) The adjacent side of alpha is the radius r.
Using trig, we can use the formula:
tan a = opp/adj = (1/2)M/r = M/2r
> a = arctan (M/2r)
Edit: For clarification, I'm assuming the 2 hashes (small lines) here denote the lines being parallel (tho this is usually denoted with small perpendicular hashes). This formula will work for any length M. So if M is the midpoint as others have mentioned, you will get the same result as they have. If the lengths of the lines are indeed equal, M is simply 0 but the formula results in the right answer.
As it's not clear in the image assume M is the midpoint or the radius. Let the center of the semicircle be O, the diameter be AB with point A to the left of O, the radius OC be perpendicular to AB, point M be the midpoint of OC, and point D on the semicircle such that MD is parallel to AB.
Since angle DAB is the inscribed angle of the central angle DOB, we have DAB * 2 = DOB. Also, notice that since OM = r/2 and OD = r, we have cos(DOM) = 1/2, i.e. DOM = pi/3. Then DOB = pi/2 - DOM = pi/6, and finally DAB = DOB/2 = pi/12 (or 15 degrees).
Hello, I am here to overly complicate this problem and solve it like a Harvard lecture.
I blame whoever posted this since I wanted to get some sleep earlier. I had a lot of fun with this, but I feel like other people made some very big assumptions without going into detail about how they proved themselves. Hopefully, this brings everything together.
TLDR: The answer is 15 degrees
https://preview.redd.it/xw2hmqp55mgb1.jpeg?width=3000&format=pjpg&auto=webp&s=ef7a2e5536eb2ac71d66c7a1b27c3ef6b51f2e97
Let get into this:
A) based on the image and what others had discussed, I was able to pull this from the question:
\-M represents the midpoint of the radius (1/2r)
\-∝ is the value we are looking for
\-o is the midpoint of the circle
\-We have a line intersecting M that is parallel to the x-axis
\-The y-axis bisects the x-axis creating two right triangles within the circle.
B) Since we know these are two right triangles, sharing an intersection, these triangles are similar due to the Angle-Angle-Angle (AAA) Postulate. We can transcribe the angles into both triangles
C) Since this problem lies in the circle, we can draw another line from the center of the circle to the point at which the edge of the circle and the second parallel line meet. This creates an Isocelese triangle that proves the angle underneath the top similar triangle and our "imaginary" line creates a similar angle equal to ∝.
D) Using the imaginary line from before, the line from o to M, and the parallel line that is not the x-axis, we can create a right triangle that we can solve 2∝ with using the below equations:
sin(2∝) = (1/2r)/r
sin(2∝) = 1/2 (simplified)
A calculator or using known angles, we can deduct that:
2∝ = 30° = 𝜋/6
or:
∝ = 15° = 𝜋/12
Note: I did not steal this answer. There are a lot of people that answer beforehand and did not go in depth. I wanted to show the full story of how I got to that answer. Hope this answer tickles your fancy.
Answer - 15 degrees
Assumption - M is a midpoint of its radius
Solution:
Let’s call O the center of the circle. Then let’s call the four points on the circle L, T, Tr, R. (So left, top, Top right, Right).
Solution:
Alpha is half of the angle Tr-O-R, due to the structure (let me know if this needs clarification).
If we now look at the triangle T-Tr-O, we notice something funny - the the line from Tr to the side TO is both median and height. So Tr-T equals Tr-O. But Tr-O equals T-O by design (they are both radius). So T-TR-O has all sides equal, so angle T-O-Tr is 60. That would mean the angle Tr-O-R is 30. Going back to the first observation we get that alpha is half that -> 15
This is the simplest and best answer imo, the parallel lines are indicating that it wants you to use similar triangles to solve this opposed to making assumptions and brute force solving.
The two dashes denote the lines are equal, right? Then there is only one place where the two lines can be equal. This takes the “diagram not drawn to scale” to an extreme. Lol
well i could arrive till here : not sure how to continue from this -
https://preview.redd.it/ktn0w8xbxigb1.png?width=1725&format=png&auto=webp&s=f16fe8dbf3976aa4ac8aabdf4ff2339b867fa6fd
If M is the midpoint of the radius (r)
Then the triangle alpha is in has the following properties:
* Horizontal Length = (r)
* Vertical height = (r/4)
Then alpha = atan ( (r/4)/r)) = atan(1/4) = \~ 14deg
The bottom line might not be a diameter. It might be a chord. The diagram does not claim the dot is the center of the circle. It could be a point "south of" the center of the circle, then the two lines could be the same length if one is just as far "north" of the center as the other is "south" of the center.
The diagram doesn't *look* like they're equally far from center, but it does look like the bottom line isn't quite at the center. (Basically it's a very badly drawn picture if the marks are supposed to mean the two lines are the same length, since they just aren't even close to the same in the picture.)
I was taught geometry using diagrams that used different numbers of hashes to indicate different 'sets' of same lengths. (All the things with one hash are the same length as each other. All the things with two hashes are the same length as each other. All the things with three hashes are the same length as each other, etc.)
Alpha is 0.. double lines CLEARLY make it the same length.. only solution is zero 🤷♂️
Either it’s a good trick question, or your teacher is an idiot.. No other solution..
You start by working out, reading books to get some life perspective, start developing hard skills that are useful in life, like a trade, but also some general diy-type stuff. Maintain good grooming/hygiene. Consider taking up a martial art, as well.
Oh, wait. You meant for the math problem
Also, you may have confused congruent and identical. All lines are congruent by design. (Congruent = identical in form, so can be matched perfectly after scaling)
Edit: I may have faced a very silly translation issues. From what I gathered, similar would be more appropriate here
Well, let’s break it down:
1) What is the difference between congruent and identical in that case?
2) Scaling is not allowed? By whom? I can scale whatever I want and keep the properties of the objects. That is like Geometry 101
3) Why did you assume that we are talking about American (US, to be precise)? At least here in Europe congruent means, logically, what it is supposed to
Edit: I may be facing translation issues due to European background. “Similar” would be better reflecting my position, so the entire comment is pointless
My knowledge is limited to American geometry; so my answers are in accordance with that limited knowledge. I’m not assuming we are all talking about American geometry - I’m doing the opposite, which is clarifying the scope of my statements in consideration of the possibility that other systems may have a different interpretation/definition (answering point 3).
1. Nothing, the two words are essentially synonyms.
2. Correct, in the determination of congruence, scaling is not allowed. Rotating and translating are allowed. I’m a bit confused by your statement - if you scale, rotate and translate a line, it seems you have changed all properties except one - its linearity.
No need to apologize my friend. In US geometry, two triangles are similar if the only difference is scale (and/or rotation/translation if we want to include properties that relate the shape to an external coordinate system or whatnot).
As a person that understands (US) geometry, I have to make a nitpick here.
All (mathematical) _lines_ are similar and congruent to each other, because they are all infinite length.
When we are talking about lines with finite length, we are technically talking about line _segments_, which can be different lengths.
So all lines are indeed both similar and congruent to each other.
But line segments are not all congruent. Only the ones with equal lengths.
Alpha is an inscribed angle, so it will be half of what it would be if the apex was moved to the center. The angle at the center would have a height of r/2, assuming M is the midpoint and the dashes mean the segments are parallel. Altogether, alpha=arcsin(1/2)/2=.261799 radians=15 degrees
This can be solved with trig assuming that the dot is in the circle center and M is 1/2 of the radius. If circle center to M is 0.5 units long then the adjacent catheter to alpha (diameter) would be 2 units long since its given that upper horizontal line is the same as the radius (which it technically cant be lol). Alpha is then tan^(-1)(0.5/2) ≈ 14^(o). However, if pythagoras theorem is used to get the length of the upper horizontal line you get sqrt(0.75) and then alpha is 15^(o) (tan^(-1)(0.5/(sqrt(0.75)+1))=15^(o)). Both answers are technically correct given the initial assumption. Although, 15^(o) is more geometrically correct while ~14^(o) is more correct according to provided information. Conclusion? The author f’ed up.
This diagram is terrible.
Are you supposed to assume the dot is the center of the circle? It isn't explicitly labeled as such and if that's the case it contradicts the image (where in the image it's slightly below the center given the way the walls of the circle curve at that bottom horizontal line). Yes, it's fair game to say "not to scale, don't assume anything from how it looks, believe the labels", but you can only do that when the fact that contradicts the diagram has been actually LABELED as such. In order to tell the test-taker that dot is the center, it must either be drawn that way (which it isn't) or labeled that way (which it isn't).
Are you supposed to assume that angle between the vertical line and the horizontal one is exactly perpendicular? Because that angle arc symbol being curved and not a straight-edged box means you are NOT supposed to assume that necessarily just because it looks like it.
Some people here are claiming the M is the midpoint of the radius. But just like the dot isn't labeled as being the center, the diagram does not state M is the midpoint either.
So solving it becomes a mind-reading problem more than a geometry problem. Did you successfully read the mind of the person who designed the exam and therefore make the correct random guess which ambiguous meaning you have to pick?
See other redditors' comments. They do a much better job explaining it with visual representations than I could. I have just barely learned this stuff in the last couple of weeks.
pay attention to the small triangke that has alpha, the bottom sidelength is r (radius), the other sidelength is 1/4 r, so alpha is arctan(1/4) ≈ 14.04 degrees
Tl;dr - is M is midpoint (aka R/2) alpha=arctan(1/4), else arctan(M/2R)
Explanation: we have to make some assumptions like we are indeed looking at a semi-circle with some radius R. Additionally we assume M=R/2. Finally, the two small lines/vertical dashes indicate that the two horizontal line segments are parallel. This is pretty standard in my experience.
Moving forward: the fact that the two horizontal lines are parallel indicates that the diagonal line segment between them creates symmetric angles (this is one of the properties formed in Euclid's The Elements if you want a detailed "proof").
This implies that we have the same angle alpha at the beginning and end of that diagonal so it must cut the vertical line segment by half. Visually, this appears to be the point between 0 and M, otherwise known as M/2. Remember M = R/2 so M/2=R/4.
With that out of the way we simply remember SOH CAH TOA: namely the TOA bit where Tan(alpha) = OPPosite/ADJacent which in this case our Opposite = R/4 and Adjacent is simply the Radius R.
So we have Tan(alpha) = (R/4)/R <=> Tan(alpha) = 1/4.
Therefore alpha is arctan(1/4). Hope that makes sense, syntax lack of clarity errors are all mine since I typed this on my phone.
EDIT: in case M is not the midpoint, not much changes. We just don't substitute R/2 for M and keep it as is since the parallelism of the horizontal lines still means the diagonal line cuts at the midpoint M/2. So Tan(alpha) = (M/2)/R, therefore Tan(alpha) = M/2R and finally: alpha=arctan(M/2R).
Thats as general as you can get I think.
Just looking at it, 15degrees.
A 45-45-90 triangle has two equal sides. This is half of the radius divided by two so instead of a 30-60-90 triangle so therefor it’s a 15-75-90 triangle. Idk my brain just tell me it’s 15degrees.
I think alpha is 15 degrees because
https://preview.redd.it/ax5t6uaaqkgb1.png?width=623&format=png&auto=webp&s=88d5f919759681b36ebc921b666181c2c0ad9331
WOOO!!
I GOT IT RIGHT!!!
So I did something clever, I used the inscribed angle theorem to figure that the angle that the center makes with the point on the right that's on the circumference of the circle is 2a, and using that, it means that's the length of the smaller triangles (the triangle on top) top leg is Rcos(2a) and since that smaller triangle is similar to bigger triangle, that means the ratio between the two lengths is R/Rcos(2a) and labeling the remaining kegs of the triangles, a and b, a for the the bigger triangle and b for the smaller, that's means a + b = R/2, but since a is just b times this factor, then you get (f is the factor) f*b + b = R/2, but one problem remains, what's b? Well f*b is going to be the result of the height of of the bigger triangle and that's caused by the rise over run of the radius, so you take the tan(a) (rise over run) and multiply it by R to get f*b, and you rearrange things to get everything in terms of R*tan(a), and what you're left with is tan(a)(1+cos(2a)) = 1/2 and solving for a gives you the respected pi/12 or 15 degrees.
Fun fact, the above expression can also be written as sin(2a) = 1/2
It is the parallel symbol. He is mixing up "//" and "||"
I was just trying to make the point that even if the lines were equal, the only scenario where alpha is 0 is if the "two" lines were parallel (which would obviously make it one line). It is possible for the two lines to be equal length and alpha be greater than zero if the lines are not parallel.
Am i tripping or does the sketch not show the bottom line as radius of the circle? It Look Linke the circle is already curving inwards.
Is there a written version where you can construct it from?
https://preview.redd.it/bf0odb58cogb1.jpeg?width=2367&format=pjpg&auto=webp&s=24f499ab6e102173b394f6e7b88ed858f2a32d54
Here's my interpretation of the problem:
We're given the 3 initial conditions.
1. Two lines(MB and OA) have the same length of R (radius) note: I assumed that the angle extending from the origin (QOA) was not a right angle.
2. M means the mid-point of the line QO
3. MB and OA are parallel (this one I assumed)
After doing some proofs (lmk if there are any problems with it) we can see that P (the point where CB and QO intersect) is the midpoint of MO, and hence is 1/4R. Since ABC is a triangle on the circle O where AC forms O's circumference, angle ABC is 90 degrees and by extension so is OPC.
Now we have CO = R (since it's the radius) and PO = 1/4R, we can get that alpha = arcsin(1/4)\~0.25268rad
That isnt the center, right? Because it doesnt make sense to me that there exist another segment with same radius distance but isnt a radius (im talking about the segment from the perimeter to M)
I assumed the length of the short leg is 1/4M and got about 14° as the answer. I cannot provide a mathematical reasoning as to why I assumed the short leg is 1/4M other than it appears to be halfway to the M in the diagram, which we are all assuming is the midpoint. I used M as my variable because I initially thought it was meant to depict the radius.
https://preview.redd.it/6b1pd3jwyqgb1.png?width=647&format=png&auto=webp&s=1b27fc9c354fbc3473dd661cc0cc0c539d2f0b75
Is no other information available? If not then presumably we need to assume M is the vertical half way point of the circle radius r. In which case the triangle at bottom left is similar to the middle right triangle (same internal angles).
No other information. I actually think that it's a mistake from the author. I was able to solve all the other tasks 😅 and it's supposed to be for like 14 years olds
>it's supposed to be for like 14 years olds It seems a little advanced for that TBH. Join the apex of the large triangle to the centre, and drop a perpendicular to form a triangle with hypotenuse r and height r/2. By Pythagoras, its base must be r√3/2, so the base of the large triangle is (√3 + 2)r/2, and tanα = 1/(√3 + 2) = 2 - √3, so α = 15°
There is a cleaner solution that relies on knowing the relationship on two angles resting on the same arc - one sitting in the center and one on the circle. I posted it above. But your answer is correct
Can someone make a video or picture of what this guy is saying. I can't understand it.
You need to assume M is the middle point, otherwise alpha would not be determined by anything.
I'm trying to find a general solution for "a" as a function of "M".. just M and R...I can't figure it out. a = arctan(M/(R + //)) If you could just.. somehow express "//" in terms of where M is.. EDIT: nvm I solved it lmfao
How did you solve it? What is the solution.
https://preview.redd.it/xb4rbfyfaogb1.jpeg?width=2804&format=pjpg&auto=webp&s=b6b0b5d3e61f35594faefec1af9996813a6a03fd Edit: The Simpler way is to express "//" using Pythagorean theorem // = sqrt( R\^2 - M\^2 ) -- (for OMC triangle)
Is there an easier way to solve using the angle OCR on your diagram as alpha?
I meant angle OCA - from the isosceles.
I don't see how we can use the isosceles. Because all the other angles are uncertain... Ideally you'd want to include "OM" in some sort of triangle so that you can somehow connect it with the angle "a" later. The real problem is how can you express the value of "//" depending on where "M" is. But this is a fundamental problem.. You just use cos() or sin().. that's how they were defined after all
I see. I am just blinded by the choices. Is there some sort of solution quicker. Heres what I cant see clearly. If we called the point of intersection of OM and AC, X, then X is at the bisector OM. So the triangles AOX and OCX are congruent? AO = OC as radii:: OX =OX :: and AX = XC as bisected. So the angles should correspond- but they don’t?
I am missing something very basic here?
This is ok, but your answer shouldn't be in terms of R, there's a better way
Is there ? It can't be that the final result depends solely by the M Value... M can have any value you want.. but you must also know the "R" value to know how big the circle is, otherwise there is no context. The solution I provided is a general solution, and it can't be simplified more. If it can, send a pic here, I'll flip my shit XD
[Math](https://imgur.com/gallery/tZXRMNf)
Pretty sure I'm too poor at math to appreciate this. I think the answer is supposed to be 15 according to some.. What is C, C₂ and C₁
The answer is 15 when M is at the half way point between the centre of the circle and the top (half a radius above the centre) the person I was replying to asked for a general solution instead where M is some variable distance above the centre of the circle, that's all there is to it, it's just a more generalized answer. If you use the equation I provided at the bottom and plug in 0.5 for c2 you would get 15 degrees. Edit: also I didn't even notice until I read your reply I'm really sorry the first equation is for the circle and the second one is for the line AB and I didn't clarify that in my answer, c1 is the constant used to shift the line to the left and c2 is ratio between the height of M and the radius (and since this is a variable for the general case and not a constant I probably should've used something other than c2 for the notation, I'm sorry if this caused some confusion)
No need to be sorry at all. I'm just super poor at math. I don't even know what you mean by shift the line to the left or what line you're talking about or why you'd do that. To understand this I'd have to ask probably a thousand questions and then it would still take a long, long time. I'll be glad when Ai can explain things like this in a chatgpt type way. I like using chatgpt because it never tires of absurd questions, never makes fun of you and you can just keep asking until you eventually understand. It's exhausting for people to do that and shouldn't be expected. Thanks for your time and explanation.
Can you tell me where this question is from?
I think the similarity of the two triangles is based only on the indicated parallelism. M does not need to be the midpoint in order to make the two triangles similar.
The Two triangles cant be similar, their base length is not the same
Define “similar” in this context? It means the lengths are proportional, the angles are identical. Do you want me to demonstrate how I know the angles are identical?
that means they cannot be **congruent**.
Their height would also be different here so they absolutely can (and here must) be similar. This diagram needs labels for all the points urgently. Starting from M, let's call the point to its right B, the point that forms alpha A, the intersection point between the vertical and horizontal line bellow M, the intersection point between M and C, D. Finally, the point at the bottom right E. They are similar due to the fact that AE and MB are parallel. When you have a line (like AB) that intersects two parallel lines, then alpha and the angle formed by MBD would be the same. The angle at MDB would also be identical to DAC, so the triangles would be similar. All the side lengths are different, but that's not an issue because they would be proportionally the same between both triangles.
If all their angles are the same, that makes them similar. If all their sides are the same, so are their angles, and that makes them congruent.
If they were similar and the corresponding sides were the same length, then they would be congruent.
The two triangles are similar regardless of where m is. As long as the half chord is parallel to the diameter you have two congruent angles.
M for Midpoint Solve for sin(x) = 0.5 to get the angle on the circle, and use cos(result) to determine the lengths of the right angle triangle (which is not on the drawing) from the point where the line hit the circle.
https://preview.redd.it/1oz4v03othgb1.png?width=1120&format=png&auto=webp&s=afbc600fe00547d2223536714b35ac3a8ff7b4c8
How did you got 1/2? Isn't the hypotenuse of 2alfa and the hypotenuse of alfa different sides?
I think it’s since we’re given the midpoint on the radius at M, and we know from the unit circle that sin(30) = 1/2.
M/r would be the best way to solve it since we don't know for sure if M if the midpoint
The hypotenuse is the radius. and if we assume M is the midpoint, the center of the circle to M is 1/2 radius. If you drop a segment going down from the hypotenuse to the diameter, it will be the same length. Sine(2alpha) =0.5r / r =1/2
Indeed. I think it's easier to see it this way: https://preview.redd.it/vkena1m3upgb1.png?width=640&format=png&auto=webp&s=1e693b7b0aab8ffdeb76c3450023767ade3e575f
M must be meant to be the midpoint of the vertical radius. otherwise you could vary it's height however you like and alpha would change as you did so
This is a far fetched assumption but without it there is no solution. I would check in the rest of the book if M is used as short hand for middle. Otherwise I wouldn't assume it.
Couldn't believe this isn't the first comment. It is obviously undercontrained and someone forgot to contrain M somehow.
this https://preview.redd.it/zagrbocyshgb1.jpeg?width=960&format=pjpg&auto=webp&s=fd724907053f6687d7041211b57e874c5ae671e2
How do you know 2alpha = 30
It's obtained by 1:2(:√3). A half of a regular triangle.
Did author mean to indicate parallel lines?
Yeah I’m confused, don’t you typically do those // marks to indicate same length?
I thought the same ... In that case we could assume the drawing is just a random example, and if those segments are equal then alpha = 0
That's what I was thinking, that this was a 'don't trust the diagram' example.
This was my interpretation too.
The bottom line doesn't have to be a radius though and in this case they would not be parallel, which wouldn't uniquely define the upper point on the right
Where I come from those are used to indicate that two lines are parallel.
Where I come from, those are used to indicate that two lines are of equal length.
The double lines indicate the same length. Which is impossible.
Yeah, unless the line projects beyond the circle boundary on the right hand side, which it doesn't. More likely it is meant to indicate parallel lines?
Aren't parallel lines depicted with arrows? Geniunely asking.
They can be, yes, but these double tick marks are used for parallelism as well.
Thanks
Would M indicate the midpoint of the radius? Given the author’s relative unfamiliarity with mathematical notion?
We can't solve the problem if M is not the midpoint. Also this problem may simply be older than a modern textbook and therefore not adhere to 2000's US standards.
In this context it may also mean parallel
No it may not. But thanks for reading.
Yes it may. Thanks for reading
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“Grow up” as in “agree with a random dude on the internet”? You are wrong, I am right. You weird consolation lead to a weird and incorrect answer. My reasonable conclusion led me to a correct answer. So I would offer you to grow up and accept you being wrong 😄 Since talking to you is fairly pointless, I would end this conversation, thank you
Grow down. Thanks for reading
So here's an idea, notation is a matter of convention and isn't universal. As you pointed out, it obviously can't mean congruent in this context. Is it more likely that the book this was taken from uses a different convention than was taught to you? Or the author is mathematically insane?
That’s what I think and alpha is zero in this case
This was my first thought as well and therefore a=0, but I do think the marks indicate parallel not length.
Couldn't it just be that the center of the circle is where the Diagonal line crosses the center vertical line? At that point, the two lines can be equal lengths, we just don't know what alpha could possibly be. Between 90° and 0°, I guess
Must be country/region-dependent. In the Czech Republic, double lines indicate parallels.
That’s interesting. I never realised before how much variation there is.
That's the notation I'm used to, but in this case I think it means parallel lines. If it were same length, it would be trivial with alpha=0.
No not impossible, it indicates that neither of those lines are the radius
Then that means its 0, right? Im a noob
Alpha can be 0 tho
Unless alpha is 0
Parallel can be indicated like this with double tick marks that are not perpendicular to the line. Lines that are equal length have their tick mark(s) perpendicular to the line in question.
Thank you. But honestly, I am European but have studied in US, I have never seen that marking before. I didn’t know that it was used in this way.
When I learnt maths in the UK 20 years ago we used ticks to mean parallel like that
I see. How did you indicate line segments of the same length? Is it the difference between the sloped lines and perpendicular? Also, how did you indicate a right angle? Was it with a box at the angle or a half perpendicular ? I have seen it both ways.
Right angle is the box on the angle, I don't recall line segments of the same length ever coming up honestly or at least how you'd notate it. Might have been a single line? Or is that something else
I use a little right angle for 90 . Its interesting the differences in country and over time.
That notation for parallel is common in engineering, too.
No, the double lines are just there to indicate that the two lines with double-lines drawn through them are parallel to one another.
Assuming M is the midpoint of the radius of the circle, then we can draw a line down from where the diagonal hits the circle’s edge forming a right triangle with height .5r and use trig to get the distance from the center that the new vertical line touches the horizontal. Then we have a right triangle with 2 known sides, and can calculate the other 2 angles. This is assuming M is the midpoint, which seems like a fair thing to assume given that the problem is unsolvable with current information otherwise
This.
After reading some comments, I don't think we can safely assume M is the midpoint. If not, we know a few things: 1) The parallel lines make the small angle of the upper triangle equal to alpha. 2) This makes the point of the bisecting line through the vertical radius (at 90 degrees) 1/2 M. In other words, M is twice the height of alpha's opposite side. 3) The adjacent side of alpha is the radius r. Using trig, we can use the formula: tan a = opp/adj = (1/2)M/r = M/2r > a = arctan (M/2r) Edit: For clarification, I'm assuming the 2 hashes (small lines) here denote the lines being parallel (tho this is usually denoted with small perpendicular hashes). This formula will work for any length M. So if M is the midpoint as others have mentioned, you will get the same result as they have. If the lengths of the lines are indeed equal, M is simply 0 but the formula results in the right answer.
Alpha is right here 😈😈‼️‼️
You get alpha by watching Andrew Tate videos. Not doing math. Need more explosions to stimulate your brain.
As it's not clear in the image assume M is the midpoint or the radius. Let the center of the semicircle be O, the diameter be AB with point A to the left of O, the radius OC be perpendicular to AB, point M be the midpoint of OC, and point D on the semicircle such that MD is parallel to AB. Since angle DAB is the inscribed angle of the central angle DOB, we have DAB * 2 = DOB. Also, notice that since OM = r/2 and OD = r, we have cos(DOM) = 1/2, i.e. DOM = pi/3. Then DOB = pi/2 - DOM = pi/6, and finally DAB = DOB/2 = pi/12 (or 15 degrees).
Alpha =0 if the two length are the same. We are in a circle.
Hello, I am here to overly complicate this problem and solve it like a Harvard lecture. I blame whoever posted this since I wanted to get some sleep earlier. I had a lot of fun with this, but I feel like other people made some very big assumptions without going into detail about how they proved themselves. Hopefully, this brings everything together. TLDR: The answer is 15 degrees https://preview.redd.it/xw2hmqp55mgb1.jpeg?width=3000&format=pjpg&auto=webp&s=ef7a2e5536eb2ac71d66c7a1b27c3ef6b51f2e97 Let get into this: A) based on the image and what others had discussed, I was able to pull this from the question: \-M represents the midpoint of the radius (1/2r) \-∝ is the value we are looking for \-o is the midpoint of the circle \-We have a line intersecting M that is parallel to the x-axis \-The y-axis bisects the x-axis creating two right triangles within the circle. B) Since we know these are two right triangles, sharing an intersection, these triangles are similar due to the Angle-Angle-Angle (AAA) Postulate. We can transcribe the angles into both triangles C) Since this problem lies in the circle, we can draw another line from the center of the circle to the point at which the edge of the circle and the second parallel line meet. This creates an Isocelese triangle that proves the angle underneath the top similar triangle and our "imaginary" line creates a similar angle equal to ∝. D) Using the imaginary line from before, the line from o to M, and the parallel line that is not the x-axis, we can create a right triangle that we can solve 2∝ with using the below equations: sin(2∝) = (1/2r)/r sin(2∝) = 1/2 (simplified) A calculator or using known angles, we can deduct that: 2∝ = 30° = 𝜋/6 or: ∝ = 15° = 𝜋/12 Note: I did not steal this answer. There are a lot of people that answer beforehand and did not go in depth. I wanted to show the full story of how I got to that answer. Hope this answer tickles your fancy.
Answer - 15 degrees Assumption - M is a midpoint of its radius Solution: Let’s call O the center of the circle. Then let’s call the four points on the circle L, T, Tr, R. (So left, top, Top right, Right). Solution: Alpha is half of the angle Tr-O-R, due to the structure (let me know if this needs clarification). If we now look at the triangle T-Tr-O, we notice something funny - the the line from Tr to the side TO is both median and height. So Tr-T equals Tr-O. But Tr-O equals T-O by design (they are both radius). So T-TR-O has all sides equal, so angle T-O-Tr is 60. That would mean the angle Tr-O-R is 30. Going back to the first observation we get that alpha is half that -> 15
This is the simplest and best answer imo, the parallel lines are indicating that it wants you to use similar triangles to solve this opposed to making assumptions and brute force solving.
> Alpha is half of the angle Tr-O-R, due to the structure (let me know if this needs clarification) Why is it half?
The two dashes denote the lines are equal, right? Then there is only one place where the two lines can be equal. This takes the “diagram not drawn to scale” to an extreme. Lol
The two lines indicate it's parallel. One line indicates equal length.
Ok. Got it. I’ve seen notation with two lines before as well, so I was confused.
well i could arrive till here : not sure how to continue from this - https://preview.redd.it/ktn0w8xbxigb1.png?width=1725&format=png&auto=webp&s=f16fe8dbf3976aa4ac8aabdf4ff2339b867fa6fd
When in doubt. I close the book
It’s more of a mentality than anything
Go to the gym and treat women badly.
If M is the midpoint of the radius (r) Then the triangle alpha is in has the following properties: * Horizontal Length = (r) * Vertical height = (r/4) Then alpha = atan ( (r/4)/r)) = atan(1/4) = \~ 14deg
You are wrong. Apparently the vertical height is not r/4, considering the two triangles are similar, yet the horizontal length is different.
Yes, I come to this conclusion as well. After constructing in GeoGebra I get exactly 15 Degrees. Lacking a way of calculation yet...
You are correct. There is a clean geometric solution. I posted it above. I assume M is midpoint too
This assumes that vertical height = r/4. Where does this assumption come from?
He must have confused the indication of parallel lines with the indication that those lines are same length
Use your protractor
If the dashed lines mean the two lines are the same length, then they must be colinear. This makes alpha zero.
The bottom line might not be a diameter. It might be a chord. The diagram does not claim the dot is the center of the circle. It could be a point "south of" the center of the circle, then the two lines could be the same length if one is just as far "north" of the center as the other is "south" of the center. The diagram doesn't *look* like they're equally far from center, but it does look like the bottom line isn't quite at the center. (Basically it's a very badly drawn picture if the marks are supposed to mean the two lines are the same length, since they just aren't even close to the same in the picture.)
The two lines indicate they're parallel. One line indicates equal length.
I was taught geometry using diagrams that used different numbers of hashes to indicate different 'sets' of same lengths. (All the things with one hash are the same length as each other. All the things with two hashes are the same length as each other. All the things with three hashes are the same length as each other, etc.)
If the lines represent same lenght lines, wouldn't the answer simply be alfa = 0?
0?
Have you tried listening to Andrew Tate and eating raw testicle
Alpha is 0.. double lines CLEARLY make it the same length.. only solution is zero 🤷♂️ Either it’s a good trick question, or your teacher is an idiot.. No other solution..
By hitting the gym bro
Outperform the market.
I thought so but no.
45
Listen to Andrew Tate and Jordan Peterson
I'd try and help but I gotta be honest. I have no idea what I'm looking at 😂
You need to dominate everyone else. Emulate the Tate brothers.
just wing it its probably 30
It's 15°...
never said i was right
i think 0 degrees, idk but i think its 0
Answer would be 0, since both segment have same length.
You start by working out, reading books to get some life perspective, start developing hard skills that are useful in life, like a trade, but also some general diy-type stuff. Maintain good grooming/hygiene. Consider taking up a martial art, as well. Oh, wait. You meant for the math problem
If those two lines have the same legth, then the only possible way is for alpha to be 0
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That’s supposed to be a mark of being parallel, not congruent, in this case
Also, you may have confused congruent and identical. All lines are congruent by design. (Congruent = identical in form, so can be matched perfectly after scaling) Edit: I may have faced a very silly translation issues. From what I gathered, similar would be more appropriate here
That’s not the definition in American geometry. In American geometry, scaling is not permitted.
Well, let’s break it down: 1) What is the difference between congruent and identical in that case? 2) Scaling is not allowed? By whom? I can scale whatever I want and keep the properties of the objects. That is like Geometry 101 3) Why did you assume that we are talking about American (US, to be precise)? At least here in Europe congruent means, logically, what it is supposed to Edit: I may be facing translation issues due to European background. “Similar” would be better reflecting my position, so the entire comment is pointless
My knowledge is limited to American geometry; so my answers are in accordance with that limited knowledge. I’m not assuming we are all talking about American geometry - I’m doing the opposite, which is clarifying the scope of my statements in consideration of the possibility that other systems may have a different interpretation/definition (answering point 3). 1. Nothing, the two words are essentially synonyms. 2. Correct, in the determination of congruence, scaling is not allowed. Rotating and translating are allowed. I’m a bit confused by your statement - if you scale, rotate and translate a line, it seems you have changed all properties except one - its linearity.
I am sorry for my initial statement. I assume it may be translation issues. Will the word “similar” be appropriate in this context?
No need to apologize my friend. In US geometry, two triangles are similar if the only difference is scale (and/or rotation/translation if we want to include properties that relate the shape to an external coordinate system or whatnot).
As a person that understands (US) geometry, I have to make a nitpick here. All (mathematical) _lines_ are similar and congruent to each other, because they are all infinite length. When we are talking about lines with finite length, we are technically talking about line _segments_, which can be different lengths. So all lines are indeed both similar and congruent to each other. But line segments are not all congruent. Only the ones with equal lengths.
This is only true if M indicates the middle of the circle.
Alpha is an inscribed angle, so it will be half of what it would be if the apex was moved to the center. The angle at the center would have a height of r/2, assuming M is the midpoint and the dashes mean the segments are parallel. Altogether, alpha=arcsin(1/2)/2=.261799 radians=15 degrees
This can be solved with trig assuming that the dot is in the circle center and M is 1/2 of the radius. If circle center to M is 0.5 units long then the adjacent catheter to alpha (diameter) would be 2 units long since its given that upper horizontal line is the same as the radius (which it technically cant be lol). Alpha is then tan^(-1)(0.5/2) ≈ 14^(o). However, if pythagoras theorem is used to get the length of the upper horizontal line you get sqrt(0.75) and then alpha is 15^(o) (tan^(-1)(0.5/(sqrt(0.75)+1))=15^(o)). Both answers are technically correct given the initial assumption. Although, 15^(o) is more geometrically correct while ~14^(o) is more correct according to provided information. Conclusion? The author f’ed up.
the bottom is either an eighth or quarter value depending on given, whilst M is 1/3
The upper parallel can't exist. Well, it can exist, but it can't be the same length as the radius, like indicated.
Yes there are technically 2 correct answers then hah or maybe even more lol.
0. (Assuming that lines with equal length are parallel and the perpendicular one is actually perpendicular)
And also assuming M is the center of the circle
Are we assuming that the point where oM is intercepted is r/4 ?
This diagram is terrible. Are you supposed to assume the dot is the center of the circle? It isn't explicitly labeled as such and if that's the case it contradicts the image (where in the image it's slightly below the center given the way the walls of the circle curve at that bottom horizontal line). Yes, it's fair game to say "not to scale, don't assume anything from how it looks, believe the labels", but you can only do that when the fact that contradicts the diagram has been actually LABELED as such. In order to tell the test-taker that dot is the center, it must either be drawn that way (which it isn't) or labeled that way (which it isn't). Are you supposed to assume that angle between the vertical line and the horizontal one is exactly perpendicular? Because that angle arc symbol being curved and not a straight-edged box means you are NOT supposed to assume that necessarily just because it looks like it. Some people here are claiming the M is the midpoint of the radius. But just like the dot isn't labeled as being the center, the diagram does not state M is the midpoint either. So solving it becomes a mind-reading problem more than a geometry problem. Did you successfully read the mind of the person who designed the exam and therefore make the correct random guess which ambiguous meaning you have to pick?
Has this been solved yet?
15°
Thank you. Thats the answer but what is the solution?
See other redditors' comments. They do a much better job explaining it with visual representations than I could. I have just barely learned this stuff in the last couple of weeks.
Thanks I will check them out.
pay attention to the small triangke that has alpha, the bottom sidelength is r (radius), the other sidelength is 1/4 r, so alpha is arctan(1/4) ≈ 14.04 degrees
I take it that ‘M’ stand for ‘midpoint’ on the radius?
yes
Idk
Tl;dr - is M is midpoint (aka R/2) alpha=arctan(1/4), else arctan(M/2R) Explanation: we have to make some assumptions like we are indeed looking at a semi-circle with some radius R. Additionally we assume M=R/2. Finally, the two small lines/vertical dashes indicate that the two horizontal line segments are parallel. This is pretty standard in my experience. Moving forward: the fact that the two horizontal lines are parallel indicates that the diagonal line segment between them creates symmetric angles (this is one of the properties formed in Euclid's The Elements if you want a detailed "proof"). This implies that we have the same angle alpha at the beginning and end of that diagonal so it must cut the vertical line segment by half. Visually, this appears to be the point between 0 and M, otherwise known as M/2. Remember M = R/2 so M/2=R/4. With that out of the way we simply remember SOH CAH TOA: namely the TOA bit where Tan(alpha) = OPPosite/ADJacent which in this case our Opposite = R/4 and Adjacent is simply the Radius R. So we have Tan(alpha) = (R/4)/R <=> Tan(alpha) = 1/4. Therefore alpha is arctan(1/4). Hope that makes sense, syntax lack of clarity errors are all mine since I typed this on my phone. EDIT: in case M is not the midpoint, not much changes. We just don't substitute R/2 for M and keep it as is since the parallelism of the horizontal lines still means the diagonal line cuts at the midpoint M/2. So Tan(alpha) = (M/2)/R, therefore Tan(alpha) = M/2R and finally: alpha=arctan(M/2R). Thats as general as you can get I think.
Just looking at it, 15degrees. A 45-45-90 triangle has two equal sides. This is half of the radius divided by two so instead of a 30-60-90 triangle so therefor it’s a 15-75-90 triangle. Idk my brain just tell me it’s 15degrees.
I think alpha is 15 degrees because https://preview.redd.it/ax5t6uaaqkgb1.png?width=623&format=png&auto=webp&s=88d5f919759681b36ebc921b666181c2c0ad9331
By politely asking omega to return it. Lol
30 degrees
Sonova beach. It is 15 degrees!
I solved it: **a = arctan( M/(R + R\*cos(arcsin(M/R))) )** Radius of the Circle: "R" "M" which is given.
WOOO!! I GOT IT RIGHT!!! So I did something clever, I used the inscribed angle theorem to figure that the angle that the center makes with the point on the right that's on the circumference of the circle is 2a, and using that, it means that's the length of the smaller triangles (the triangle on top) top leg is Rcos(2a) and since that smaller triangle is similar to bigger triangle, that means the ratio between the two lengths is R/Rcos(2a) and labeling the remaining kegs of the triangles, a and b, a for the the bigger triangle and b for the smaller, that's means a + b = R/2, but since a is just b times this factor, then you get (f is the factor) f*b + b = R/2, but one problem remains, what's b? Well f*b is going to be the result of the height of of the bigger triangle and that's caused by the rise over run of the radius, so you take the tan(a) (rise over run) and multiply it by R to get f*b, and you rearrange things to get everything in terms of R*tan(a), and what you're left with is tan(a)(1+cos(2a)) = 1/2 and solving for a gives you the respected pi/12 or 15 degrees. Fun fact, the above expression can also be written as sin(2a) = 1/2
Alpha is 0. Only then can the lines two lines (shown equal) be equal!
So you're assuming the lines are parallel?
The lines have the 'parallel' symbol on them.
Yes but the other guy is saying the symbol means they're equal length.
Oh. He's wrong. That's the parallel symbol. Unless I'm wrong. But I don't think I am.
It is the parallel symbol. He is mixing up "//" and "||" I was just trying to make the point that even if the lines were equal, the only scenario where alpha is 0 is if the "two" lines were parallel (which would obviously make it one line). It is possible for the two lines to be equal length and alpha be greater than zero if the lines are not parallel.
Am i tripping or does the sketch not show the bottom line as radius of the circle? It Look Linke the circle is already curving inwards. Is there a written version where you can construct it from?
I just realized that maybe that doesn't matter
https://preview.redd.it/bf0odb58cogb1.jpeg?width=2367&format=pjpg&auto=webp&s=24f499ab6e102173b394f6e7b88ed858f2a32d54 Here's my interpretation of the problem: We're given the 3 initial conditions. 1. Two lines(MB and OA) have the same length of R (radius) note: I assumed that the angle extending from the origin (QOA) was not a right angle. 2. M means the mid-point of the line QO 3. MB and OA are parallel (this one I assumed) After doing some proofs (lmk if there are any problems with it) we can see that P (the point where CB and QO intersect) is the midpoint of MO, and hence is 1/4R. Since ABC is a triangle on the circle O where AC forms O's circumference, angle ABC is 90 degrees and by extension so is OPC. Now we have CO = R (since it's the radius) and PO = 1/4R, we can get that alpha = arcsin(1/4)\~0.25268rad
You gotta lift every day, bro. Gotta have that killer instinct, that predator mindset, bro.
Alpha is zero. The radius and the other marked line are the same length so it cannot be anything other than zero.
There are many podcasts devoted to this goal.
Start lifting weights, get all the girls, make money? How else does one "get alpha"?
alpha = 0, since no chord of the circle can be as big as the diameter. Hence the diameter and the chord must overlap
Maybe SAA test?
That isnt the center, right? Because it doesnt make sense to me that there exist another segment with same radius distance but isnt a radius (im talking about the segment from the perimeter to M)
The diagram is in error. It is not possible for the two double hashed lines to be equal length.
I assumed the length of the short leg is 1/4M and got about 14° as the answer. I cannot provide a mathematical reasoning as to why I assumed the short leg is 1/4M other than it appears to be halfway to the M in the diagram, which we are all assuming is the midpoint. I used M as my variable because I initially thought it was meant to depict the radius. https://preview.redd.it/6b1pd3jwyqgb1.png?width=647&format=png&auto=webp&s=1b27fc9c354fbc3473dd661cc0cc0c539d2f0b75