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Grade 10, angle problem.

  1. Dec 16, 2006 #1
    grade 10, angle problem (geometry).

    1. The problem statement, all variables and given/known data

    find values of angles a, b and c in order
    [​IMG]

    the two lines withthe arrows drawn on them are parallel
    the center is at where the 94 degree angle forms

    2. Relevant equations
    none
    side note though: this is from a text book for grade 10 students in a section under the angluar rules (like in the diagram, you can tell that angle b and angle c are the same because those two lines are parallel)

    3. The attempt at a solution

    okay, i tried but i didnt get too far because i kept sending myself into loops and winding where everything cancelled out to be something like 180 = 180.

    i based my work off of the general rules for working with angles:
    a + b + 94 = 180
    a = 86 - b
    b = c

    from there i tried extending the line from a through the center (which is at the 94 degree angle) and intersecting it with the parallel line coming from the angle c. i got an equation for that but it was the exact same as the very first equation i had so subbing in b = c, gave me no values for any of the angles.

    then from the extended line i reied looking at the quadraliteral formed and using the "all interior angles in a quaderalideral sum to give 360" rule but that gave me the sum of:
    [180 - b] + c + [180 - 94] + 94 = 360 (angles reading left to right)
    but subbing in b = c, again cancelled out all the variables leaving me with no answer once more.

    i tried drawing a third chord (from a across to the other side of the line coming from angle c) to form a triangle but that led me nowhere also.

    i think im thinking too deeply into to, it might not need the that much thought, but i cant see it any other way. (i even tried drawing in a line to half the "a b 94" triangle, it didnt work either)

    im actually a little depressed now, because my little sister was the one who asked me to help her with that problem >< and i couldnt do it, now i have to ask for help *sigh* i've been looking at it for ages now but i cant see how its possible to find the angles with only that much information.
    there may be a rule for it but i couldnt find one in the text book and there were no answers because it was a challenge problem.

    if anyone could help me out here i would really appreciate it. thanks in advance.
     
    Last edited: Dec 16, 2006
  2. jcsd
  3. Dec 16, 2006 #2
    Are you sure you have copied the question down correctly ?

    From what I see the only given data in the question is the angle of 94 degrees right. This information seems completely insufficient.

    Imagine you are drawig this fig, starting from the circle. You draw the circle " of any radius " and draw two rays from the centre making an angle of 94 degrees with each other. Now choose one point from one of these rays on the circle's circumference and the other point can be chosen anywhere on the other chord and join the two points forming angles a and b. From this, is it not quite clear that the values for and b vary with the choice of the undistinct point on the chord ?
    As for the other line, it is just a parallel line drawn intersecting the circle from the line joining the two points, and as a result a=b as you noted.

    If you still have doubts try drawing the figure once again.

    Perhaps b is formed on the circumference. That would make things a whole lot easier wouldn't it ? ;)

    Regards
    Arun
     
  4. Dec 16, 2006 #3

    AlephZero

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    Your picture looks strange. They way you have drawn it, there doesn't seem to be any reason for drawing the circle.

    Check what the question says. For eample, Is angle "b" supposed to be on the circumference of the circle? Is the 94 degree angle supposed to be at the centre?
     
  5. Dec 16, 2006 #4
    Thanks for the responces ^^ I'm glad to know I'm not the only one who thinks the question is really strange. =]

    I took another look at the original question and it's the same. The angle b forms on the radius within the circle and not on the circumference, the drawing was given in the text book, I had to redraw it because the drawing in the text was a little too close to the spine of the book to scan properly, but thats how it is.

    Although I do wish the b would form on the circumference making the question so much easier. At first I thought it was a error inthe book but the question has a label saying it's very hard one, if it were the case that the b fell on the circumference then it would turn it into a very very easy one ^^; although after looking at the problem the only text book error is if they chosea problem which couldnt be solved by accident. I had thought they might have left out an angle but all the other problems in that section all have and unknown, a, b and c, and a single given value.

    There was no text with the question, there were alot of drawings like this (except for the rest were alot easier to solve) for that one question in the first line, I added the line about the parallel part incase the arrows I drew looked crooked.

    As for the centre, that is where the 94 degree angle is formed. I think I mentioned it in the problem somewhere, I can't remember ^^; but i'll put it in the problem description.

    The topic of the section was rules for angles around and within circles (like a radius is perpendicular to a tangent touching at the end of it, or a radius bisects a chord when they are perpendicular, etc..)

    Sorry for writing so much, all I can think about is the problem now and it's 1:30am. I think I'll go to bed and try it in the morning so I have a fresh mind.
     
  6. Dec 16, 2006 #5
    Well, it would be easier if the textbook atleast gave you a little more information.

    Now, by looking at the picture, I could infer that Angles A and B are equal. Yet, we can't say that, since there is no proof of it. If there were numbers in the book, please tell us.

    I would suggest to use Trig. But, you probably haven't learned that, and that there isn't enough given information.

    Are you certainly positive that there is no lengths for anything?
     
  7. Dec 17, 2006 #6
    How so ? For the angles to be equal, B would have to lie on the circumference.

    To the OP, since this is under the challenging problems section, perhaps they wanted the solver to see that there is no distinct solution.
     
  8. Dec 18, 2006 #7
    I'm still not making any progress on that problem. =\

    @carbz: I've learned trig already, I'm studying in Uni right now (but like you said, there's not enough info to use trig). The frustrating part is that the problem comes out of a grade 10 book and I'm having so much trouble with it. I don't think the text book has covered the section on Trig yet too. I've gone throught all the rules for geometry/angles I can think of, as well as all the ones in the chapter where the problem is found and I still can't find any relevent to the drawing.

    @arunbg: I really hope that was the solution, then I could stop worrying about it. I get stuck into problems for ages, makes my head hurt when I think too much ><;
     
  9. Dec 18, 2006 #8
    there really is not enough information on the problem. Let B be the pointy end of angle b, if B is on the circle, you get a=b=c=(180-94)/2=43

    obviously, you can draw another line from A (the pointy end of a) to the other two lines, and the angle b will undoubtedly change.
     
  10. Dec 18, 2006 #9
    You've already solved the problem. angle b=a since they subtend an angle of 94 at the center (say O), which makes OA=OB=radius of the circle (hence a=b). Now, if you extend A) to intersect the parallel line (say at D), it forms an angle of 94. Since, a=43, angle ADC=94, c=43. Hence Solved.
     
  11. Dec 18, 2006 #10

    cristo

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    b≠a, since the angle b is not on the circumference of the circle.
     
  12. Dec 18, 2006 #11
    Angles b&c are equal because the lines are parralel, 45.52deg, and a is 40.48.
     
  13. Dec 18, 2006 #12

    cristo

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    How've you worked that out?
     
  14. Dec 18, 2006 #13
    Sorry the angles based on my drawing, maybe diff.
     
  15. Dec 18, 2006 #14

    cristo

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    Oh, ok.. this has been puzzling me since the OP wrote the question! In my eyes it can't be answered!
     
  16. Dec 18, 2006 #15
    How could a = b? If you are basing this off the drawing, you shouldn't do that since there is no proof of anything. The only thing you could say is that b = c. The length of a is the radius, but the length of b is not close to it at all. Unless there is more information given, you can't solve it.

    Yura - Did the book ever give an answer to the problem? Or, did your teacher say the answer?
     
  17. Dec 20, 2006 #16
    You could draw a line from the parallel line at the 94 degree point, we'll call it 'F', to the other parallel line to a point we'll call 'E', and you get two similar triangles ECA and FCA. I can't get any further than that. But no side lengths are given, so you can't use trigonometry, vectors, or anything. Oh, and b = c, of course.
     
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