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Finding the Angles

  1. Jul 10, 2008 #1
    Just wandering, I'm about to begin Cal III, I don't ever remember seeing a problem like this. I found the problem in a tool and die mathematics question book with no explanation of finding the answer. I'm wandering if anyone might be able to help me see how the problem should work. In high school I never took geometry and in college I could only remember being given sides and angles, not just angles to find other angles when a circle is involved.

    I've uploaded the problem and the answers are going to be angles 1-10. How do I find the other angles?


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  2. jcsd
  3. Jul 11, 2008 #2


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    that's hardly Calculus III! More like basic geometry. The formula you need is that:
    If angle X, with vertex outside circle O, cuts the circle in two arcs, AB and CD, then the measure of angle X is (1/2)(CD- AB) where, of course, CD is the larger arc (and so farther from the vertex of angle X).
  4. Jul 11, 2008 #3
    He said he's "about to begin" Calc III.
  5. Jul 12, 2008 #4
    So, would that make angle one = 8 deg?

    I took angle one=.5(156-140)=8 deg. Then angle 3 is equal to angle 1, right?
  6. Jul 12, 2008 #5
    Or would angle one and angle three be = 4 deg? Since I only need half of the x=1/2(CD-AB) for angle one and angle three?
  7. Jul 12, 2008 #6


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    "I took angle one=.5(156-140)=8 deg. Then angle 3 is equal to angle 1, right?"

    No, angle 1 is 66 degrees.
  8. Jul 12, 2008 #7
    I don't even think the theorem listed above is needed, though it is important. Anyways if you've never taken geometry before, you might need a few more facts to solve this problem:

    1. A central angle is an angle whose two sides are radii of the circle. Its measure is equal to that of the arc it intercepts. Also, if we're given the measure of the central angle, then of course the arc measure is equal to it.

    2. An inscribed angle is an angle whose two sides are chords (a chord is simply a segment whose endpoints are on the circle, e.g. the diameter) that meet at a common vertex on the circle. Its measure is equal to 1/2 the arc it intercepts. Can you find angle 7 now? Also, angle BDE is VERY IMPORTANT. It's measure is equal to half of arc BFE, but arc BFE is cut off by a central angle that is a diameter, and therefore a straight line. It follows that angle BDE is a right angle. Finding right angles is VERY useful.

    3. An angle formed by a tangent and a chord is equal to 1/2 the arc it cuts off. What happens when the chord is a diameter? Do you see why angles PAO and PEO (angle 4) are right angles? Can you find angle 8 now?

    4. An angle formed by two chords is equal to 1/2 the sum of the measures of the intercepted arcs. Can you find angle 9 now?

    Okay, so armed with those facts, you should be able to find angle 7, 4, 8, and 9. Now let's proceed with some angle chasing. First of all, two intersecting chords form vertical angles. Can you find the angle across from angle 9? Now can you find angle 6? Since we know triangle BDE is a right triangle and we know angle 7, can we find angle BED? Can you find angle 5 now (did you find 6)? How about angle 10?

    Now we have angles 1, 2, and 3 left. The previous suggestion was to use theorems relating the outside vertex to arcs of the circle. Instead, we can proceed to show that angle 1 is equal to angle 3. Remember from above that PAO and PEO are right angles? There is a theorem that states that if the hypotenuse and one leg of a right triangle equal that of another, the triangles are congruent. Now consider right triangles PAO and PEO. PO = PO and note that OA = OE (why?). It follows that these two triangles are congruent, so knowing what you know about central angles, find angle 2, then find angles 1 and 3.

    Alternatively, from a general theorem known as power of a point, there is a case that states that two tangents from a point to a circle are always equal. Thus PA = PE, from which it follows that all three sides of PAO are equal to that of PEO and thus the two triangles are obviously congruent.
  9. Jul 12, 2008 #8
    No, I think angle 1 is 12 degrees. Please don't post answers, especially if you don't show any explanation.
  10. Jul 12, 2008 #9
    So Snipez90,

    Angle seven= 30 deg, because the arc it intercepts, ED, is 60 deg? Which with the right angle, the three angles of BDE, are 30 deg, 60 deg, and 90 deg?
  11. Jul 12, 2008 #10
    Yes, from arc ED we get angle 7 = 30 degrees. There is a more clever way. Draw in OD. Can you use this to show that triangle BED is indeed a 30-60-90 degree triangle?

    Point O is the circumcenter, which is the intersection of perpendicular bisectors of right triangle BED. Since a chord that passes through the center of the circle is the diameter and is a straight angle, the arc it intercepts must be 180 deg. Hence, BDE must be (1/2)(180) = 90 degrees.

    Another related key idea is that given a triangle, if a median is drawn to the opposite side and is equal to one half that side, the triangle must be right. Can you show this by convincing yourself that a circle must pass through each of the vertices of the triangle?
  12. Jul 13, 2008 #11


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    Listen snipez, I'll decide how I wish to answer any particular question as I wish. In this case I wanted to point out to OP that his answer was incorrent and allow him to have another crack at it.

    BTW. You're wrong too. Angle number one is 66 degress.
  13. Jul 13, 2008 #12
    angle 1 = 90 - angle 2 = 90 - (156/2) = 12

    This may be incorrect, but if the drawing is to scale then angle 1 is not 66 degrees.
  14. Jul 13, 2008 #13


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    This type of question is seldom drawn to scale jim, they don't want you to solve it with a protractor. Just look at angle AFE for example, in the figure it looks more like 100 to 110 degrees than 154 degrees, but it's given as 154 degrees so that's what we use.

    The problem there is that angle_2 is not (156/2) degrees. Check your working and you'll find that it's actually (180-156)/2 degrees. It must be the "not to scaleness" that's throwing everybody off the scent.
    Last edited: Jul 13, 2008
  15. Jul 13, 2008 #14
    Wow uart. It sure helps a lot to point out an "incorrent" answer and offer another incorrect answer. Regardless of whether your answer was correct or not, including no explanation WHATSOEVER doesn't help someone who's not too familiar with geometry.

    (180-156)/2 gets you half of arc AB, congrats, it doesn't help you solve the problem at all. I doubt you read any of what I wrote, but I'll prove you incorrectly in other ways. First of all, if you wanted to apply HallOfIvy's suggestion, you would do better to apply it correctly.

    "If angle X, with vertex outside circle O, cuts the circle in two arcs, AB and CD, then the measure of angle X is (1/2)(CD- AB) where, of course, CD is the larger arc (and so farther from the vertex of angle X)."

    Vertex P cuts the circle into two arcs: ACE (larger arc) and AFE, which make up 360 degrees. We're given AFE = 156 so ACE = 360-156 = 204. Now you can apply the theorem, Angle P = (1/2)(ACE - AFE) = (1/2)(204-156) = (1/2)(48) = 24 degrees. If angle 1 + angle 3 = 24, then angle 1 can't be 66 degrees.

    Don't believe the last paragraph? Consider quadrilateral PAOE. The sum of the angles is obviously 360 degrees. PAO and PEO are right angles so they add up to 180 degrees. Angle AOE is equal to the measure of AFE (central angle theorem) which is 156. Thus, the remaining angle P = 360 - (180 + 156) = 24 degrees.

    I can show you the congruency of triangles PAO and PEO (which would imply angle 1 = angle 3 = 24/2) in about 4 or 5 ways. But without even doing that I have shown your answer to be unreasonable.
  16. Jul 13, 2008 #15
    AFC is 156 degrees. The symmetry of the construction indicates that angle_2 is half that. (78 degrees).
    Since OAP is a right triangle, angle_1 is 90-Angle_2. (12 degrees)
  17. Jul 13, 2008 #16
    Correct above post to read "AFE is ...."
  18. Jul 14, 2008 #17


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    OK Jim it looks like I'm reading the angles differently. I interpreted angle AFE as representing the angle at point F when you construct lines AF and FE.
  19. Jul 14, 2008 #18
    This type of problem can be solved mechanically by drawing as many triangles as needed so that everything is triangulated, then setting up a linear equation for each triangle (sum of angles in a triangle = 180), and setting up a linear equation for each vertex (sum of angles around a vertex = 360). At this point you will have a (probably overdetermined) system of linear equations which can be solved to find all the angles.

    If there is distance information as well (actually here there is because of the radius), then you need to convert that to angle information with the law of cosines before setting up your equations.
  20. Jul 14, 2008 #19
    I should have said arc AFE. It's clear in the thumbnail sketch that its arc AFE, arc CDE and arc ED that they are talking about because they use the "hat" symbol above the letters. In any future posts I'll make sure I'm clear about whether its angles or arcs I'm talking about.
  21. Jul 14, 2008 #20
    Thanks to everyone for all of the help!

    Here is what I have gotten, hoping that I've gotten them correct!

    Angle 1= 12 deg
    Angle 2= 78 deg
    Angle 3= 12 deg
    Angle 4= 90 deg
    Angle 5= 30 deg
    Angle 6= 30 deg
    Angle 7= 30 deg
    Angle 8= 30 deg
    Angle 9= 40 deg
    Angle 10= 60 deg

    What do you guys think?

  22. Jul 14, 2008 #21
    The cut and paste did'nt work too well, but if you let the small square be the symbol
    for angle, then my results are:


    1,2, 3, 4, 5, 6, 7, 8, 9, 10
    12, 78, 12, 90, 40, 20, 30, 30, 130, 50

    Where afe, cde, and afe are the arcs
  23. Jul 14, 2008 #22
    Thanks so much!

    You guys have all been a great help. This is been some great information. I'd like to find a geometry class to go to now, just to get good. It looks like all of these theorems would be very useful in trig, not to mention real situations where this information could come in handy. It is very frustrating to know that I've never seen a problem like this, pre-cal trig or any math that I've taken at the college level.

    I suppose that this is more high school level geometry than college level, even though I'm out of high school and in college...looks like I missed out on some good knowledge.

    Thanks to everyone for helping me with this problem.
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