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Thread: Can this problem be solved? Calculating angles of triangle

  1. #1
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    Can this problem be solved? Calculating angles of triangle

    So my teacher gave this math problem to us so we can solve it, he said that it is very hard even for him to do it.The person who solves it till tomorrow will get 2x A (double A's, double 5).I can't do it, i tried everything but it just seems impossible. I'm 17 y.o. , third grade in high school.I hope someone can help me with this.


    -->We need to find those two angles that are marked on the picture, everything we know is marked and numbers on the picture are angles in degrees.Can you solve this and tell me how you get those answers? I need this ASAP. Thanks in advance! kec.jpg
    Last edited by Ariel4; 11-25-2017 at 06:20 AM.

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    I'm pretty sure extra credit requires extra effort, not a handout. Let's see your best efforts.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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    Can you answer following:

    1: what kind of triangle is ABC?

    2: angle ADB = how many degrees?

    3: angle AOE = how many degrees?
    I'm just an imagination of your figment !

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    Quote Originally Posted by Denis View Post
    Can you answer following:

    1: what kind of triangle is ABC?

    2: angle ADB = how many degrees?

    3: angle AOE = how many degrees?
    1.Isosceles triangle (with 2 equal sides)
    2.70 degrees
    3.130 degrees
    But I don't see how these answers help in any way :/

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    Quote Originally Posted by tkhunny View Post
    I'm pretty sure extra credit requires extra effort, not a handout. Let's see your best efforts.
    After 5 days of trying and consulting with another math professor both of us have no clue how to do this.I'm not even sure that this is possible to do.

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    I haven't spent time with this, but it reminds me of some similar problems I have seen, and MAY be solvable by the same methods. See here, here, and here.

    On the other hand, I drew it with GeoGebra, and the answers are not nice numbers as far as I can see; it is possible that you have a garbled version that can't be solved so easily. I'll see what I can do with it.

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    Quote Originally Posted by Ariel4 View Post
    But I don't see how these answers help in any way :/
    Just wanted to see "where you're at"...

    Change your "O" to "F" (hate using O...looks like zero!).

    Take triangle AFB:
    assign length 1 to AB; use Law of Sines:
    AF/SIN(20) = 1/SIN(130)

    Is that enough of a hint?
    I'm just an imagination of your figment !

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    You got me hooked on that darn problem!

    Here are the steps I took to solve it:

    Let AB = 1

    Calculate AF (Sine Law)
    AF = SIN(20) / SIN(130) = ~.446
    (not showing complete solution for the others;
    you do the work!)

    Calculate BF (Sine Law)

    Calculate EF (Sine Law)
    (notice that triangle AEF is isosceles)

    Calculate DF (Sine Law)

    Calculate DE (Cosine Law)

    Calculate angle DEF (Sine Law)

    angle EDF = 180 - 130 - angle DEF

    Perhaps Doc Peterson can do it in lesser steps...
    I'm just an imagination of your figment !

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    Quote Originally Posted by Denis View Post
    You got me hooked on that darn problem!

    Here are the steps I took to solve it:

    Let AB = 1

    Calculate AF (Sine Law)
    AF = SIN(20) / SIN(130) = ~.446
    (not showing complete solution for the others;
    you do the work!)

    Calculate BF (Sine Law)

    Calculate EF (Sine Law)
    (notice that triangle AEF is isosceles)

    Calculate DF (Sine Law)

    Calculate DE (Cosine Law)

    Calculate angle DEF (Sine Law)

    angle EDF = 180 - 130 - angle DEF

    Perhaps Doc Peterson can do it in lesser steps...
    Denis thank you so much for this,i solved it and got 18 and 32 degrees as solution(approximately).I gave this answer to my math teacher and he said that this is not the way i was suppose to do it. As he said apprently i don't need to use calculator and as a hint he told me to use peripheral angles and as i understood him ,try to represent angles with another angle(don't know right term for this in english but here is example. if sin130 => sin(150-20) etc and than to simplify those angles furhter more,sin and cos of double angle etc...).
    And Peterson i checked your sites also,but as far i can see it is not done in same way as my proffesor imagined(with peripheral angles).Thanks for sharing those sites tho'.

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    Quote Originally Posted by Ariel4 View Post
    Denis thank you so much for this,i solved it and got 18 and 32 degrees as solution(approximately).I gave this answer to my math teacher and he said that this is not the way i was suppose to do it. As he said apprently i don't need to use calculator and as a hint he told me to use peripheral angles and as i understood him ,try to represent angles with another angle(don't know right term for this in english but here is example. if sin130 => sin(150-20) etc and than to simplify those angles furhter more,sin and cos of double angle etc...).
    And Peterson i checked your sites also,but as far i can see it is not done in same way as my proffesor imagined(with peripheral angles).Thanks for sharing those sites tho'.
    Your answers are approximately what GeoGebra gave me, so you are presumably right; the program probably finds its answers using coordinate geometry, which is yet another possible approach.

    One of the sites I referred you to gives 12 different approaches, some geometrical, some trigonometric. There are many ways to solve any problem like this; if the teacher wanted a particular way, or a particular form of answer, he should have specified that.

    Unfortunately, I don't know what you mean by "peripheral angles", and you haven't given enough information to guess what specific approach he used.

    I myself was expecting (because of the similar problems I pointed out) that it might be able to be done exactly as integer or rational degrees, by geometrical methods rather than trigonometry. I think you are saying that his method gives exact values by using trigonometry with special angles and identities, so that you would get a radical solution. If so, then you might be able to use the same method you did, but with such exact values rather than decimals. However, since trig functions of 20 degrees are not expressible in terms of square roots (because that angle is not constructible), the best I can expect is that somehow values like sin(20) could end up canceling out, leaving you something relatively simple. Do you know at least what kind of answer he got (rational, radical, decimal, ...)?

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