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Math Help - basic geometry

  1. #1
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    basic geometry

    Please see the attachment. Could anyone tell me how to determine the value of angle x? Thanks!
    Attached Thumbnails Attached Thumbnails basic geometry-img_4693.jpg  
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  2. #2
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    Re: basic geometry

    Is there any more information given? Like are any of the segments parallel?
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  3. #3
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    Re: basic geometry

    sorry, no other info given...does that mean this problem isn't possible?
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  4. #4
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    Re: basic geometry

    Hi,
    Your problem does have a solution. Unfortunately, I can't figure out how to do it geometrically. The following uses trigonometry to find x. The answer turns out to be very close to 16 degrees. This makes me think there is some geometric solution, but I am at a loss as to what it might be.

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    Re: basic geometry

    Thanks!
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  6. #6
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    Re: basic geometry

    Quote Originally Posted by johng View Post
    Hi,
    Your problem does have a solution. Unfortunately, I can't figure out how to do it geometrically. The following uses trigonometry to find x. The answer turns out to be very close to 16 degrees. This makes me think there is some geometric solution, but I am at a loss as to what it might be.

    Looks like the angle is exactly 16 degrees.
    Based on the fact that

    \sin \left(82^{\circ }\right)-\sin \left(38^{\circ }\right)=\sin \left(22^{\circ }\right)

    x + 22 = 38

    Therefore
    x = 16
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  7. #7
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    Re: basic geometry

    Quote Originally Posted by Idea View Post
    Looks like the angle is exactly 16 degrees.
    Based on the fact that

    \sin \left(82^{\circ }\right)-\sin \left(38^{\circ }\right)=\sin \left(22^{\circ }\right)

    x + 22 = 38

    Therefore
    x = 16
    A more geometric proof of x + 22 = 38

    Construct a right triangle EDI where I = 90 degrees, DI parallel to BC, angle EDI = 22 degrees
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  8. #8
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    Re: basic geometry

    Idea,
    Could you explain to me how you get angle EDI is 22 degrees without knowing that angle x is 16 degrees? Of course angle ADI is 38.

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    Re: basic geometry

    Quote Originally Posted by johng View Post
    Idea,
    Could you explain to me how you get angle EDI is 22 degrees without knowing that angle x is 16 degrees? Of course angle ADI is 38.

    Construct EK perpendicular to BC, K on BC and as in your previous post a=BD=DE=EC

    \sin \left(82^{\circ }\right)=\text{EK}/a

    \sin \left(38^{\circ }\right)=\text{IK}/a

    \sin (\text{angleEDI})=\text{EI}/a=(\text{EK}-\text{IK})/a=\sin \left(82^{\circ }\right)-\sin \left(38^{\circ }\right)=\sin \left(22^{\circ }\right)
    Thanks from johng
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  10. #10
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    Re: basic geometry

    Hi Idea,
    Thanks for your better proof of the original OP's problem. I'm still thinking about a purely geometric solution. The following is a diagram of your proof:

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  11. #11
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    Re: basic geometry

    Quote Originally Posted by johng View Post
    Hi Idea,
    Thanks for your better proof of the original OP's problem. I'm still thinking about a purely geometric solution. The following is a diagram of your proof:

    Maybe if we can find a purely geometric proof that angle BCD = 30 degrees, it would then easily follow that angle x = 16
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  12. #12
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    Re: basic geometry

    Quote Originally Posted by Idea View Post
    Maybe if we can find a purely geometric proof that angle BCD = 30 degrees, it would then easily follow that angle x = 16
    Proof:
    Extend side BA with AF = AC so that angle BFC = 30.

    Show that triangles BCD and BFC are similar. Since they have an angle in common, this amounts to showing that BC*BC=BD*BF

    This in turn follows from

    \text{DE}^2=\text{AD}^2+\text{AE}^2-\text{AD}.\text{AE}

    and

    \text{BC}^2=\text{AB}^2+\text{AC}^2-\text{AB}.\text{AC}

    (use Pythagoras' theorem)

    Therefore angle BCD = angle BFC = 30 degrees
    Thanks from johng
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