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Thread: Please help me find two more Equation (to prove that α=β=γ=30)

  1. #1

    Please help me find two more Equation (to prove that α=β=γ=30)

    I have been trying to solve this question(please see the image)

    The question asks to prove that α=β=γ=30 .

    I am trying to solve this question by discovering at least three different Equations. As the number of unknown variables is 3.So,i must find at least 3 different Equations in order to find the values of the three different variables (α,β and γ).

    I have successfully found the first Equation

    α+β+γ=90......(i)

    Please help me find out the other two Equations!

    I will be thankful for help!

    Note: English is my second language.

    Sent from my Elite Note using Tapatalk

  2. #2
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    State the meaning of the tiny, curved marks in each angle. Do they mean we know already that all these angles are congruent?
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Quote Originally Posted by tkhunny View Post
    State the meaning of the tiny, curved marks in each angle. Do they mean we know already that all these angles are congruent?
    No,those tiny curve marks are just made to indicate the angle measure.
    They do not mean that we know that all these angles are congruent.

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    Actually there are SIX unknowns. You have not labeled every angle and considered every triangle.

    [tex]\angle AKB = \delta ,\ \angle BKC = \epsilon ,\ \angle AKC = \zeta.[/tex]

    And there are four obvious equations, one for each triangle.

    [tex]( \alpha + 30) + ( \beta + 30) + ( \gamma + 30) = 180 \implies \alpha + \beta + \gamma = 90.[/tex]

    [tex]\alpha + \delta + 30 = 180.[/tex]

    [tex]\beta + \epsilon + 30 = 180.[/tex]

    [tex]\gamma + \zeta + 30 = 180.[/tex]

    There is a less obvious equation of

    [tex]\delta + \epsilon + \zeta = \text { WHAT and why.}[/tex]

    Getting the sixth equation seems to require some information about triangle ABC or about K or about AK, BK, or CK. Have you given us all the information from the problem?

    EDIT: Depending on the geometric information given, you may not need all those equations. For example, if you are told that triangle ABC is equilateral, then you know that each angle equals 60 degrees.
    Last edited by JeffM; 12-03-2017 at 12:23 PM.

  5. #5
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    My expectation would be that you will need at least one equation involving side lengths, which means trigonometry. I say this because angles alone can't ensure that lines meet at a given place, which will be necessary if you try to construct the figure.

    Does the source of the problem imply that you should use only synthetic geometry methods, or that trigonometry would be needed? I assume you stated the entire problem as given to you, and didn't omit anything.

    Another thing to observe is that what you are told to prove is the "obvious" solution; so the problem becomes one of showing that it is the only solution. You might be able to start with that (an equilateral triangle) and show that any deviation from it will be impossible.

  6. #6
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    We could be severely overthinking this. How about [tex]\alpha = 30\;and\;\beta = 30[/tex]? Seems like two equations to me.

    I'm just not real clear on the nature of the assignment if we start with what is to be proven. Rather like Petitio Principii, I think.
    "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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    Draw 3 lines from K, one perpendicular to AB, another to BC and the third to AC. Can we calculate the angles of all 6 triangles?
    Last edited by Jonathan; 12-03-2017 at 03:06 PM.

  8. #8
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    Quote Originally Posted by Jonathan View Post
    Draw 3 lines from K, one perpendicular to AB, another to BC and the third to AC. Can we calculate the angles of all 6 triangles?
    No, because that just gives us more unknowns, 12 I suspect without doing the work.

    The whole problem is silly because incomplete. If, and only if, the three unknown angles are each 30 degrees, then the triangle is equilateral. But, so far, we have been given no information that allows us to make either conclusion.

  9. #9
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    Quote Originally Posted by navneet9431 View Post
    No,those tiny curve marks are just made to indicate the angle measure.
    They do not mean that we know that all these angles are congruent.

    Sent from my Elite Note using Tapatalk
    Sorry. Marking all six angles with the same marking MEANS all six angles are congruent. If you don't want it to mean that, you will have to use some other markings.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  10. #10
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    Quote Originally Posted by JeffM View Post
    No, because that just gives us more unknowns, 12 I suspect without doing the work.
    Not really because it is easy to see the value of all newly created angles in terms of a, b and c (I can't do the Greek characters!).
    I think there is enough information, drawing the figure seems to show that triangle ABC is equilateral. I think any proof will demonstrate that the three angles are equal using similar triangles showing that there is a symmetry to the figure. Whether this would be regarded as a rigorous proof, I don't know.

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