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Math Help - Find measures of angles, massive error !

  1. #1
    Newbie Monocerotis's Avatar
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    Find measures of angles, massive error !

    So this is the triangle I'm dealing with.

    I have to find the measures of the angles x & y.

    First thing I tried to do was find the measure of the angle @ A.

    a/SinA = c/SinC
    66/SinA = 25/Sin10.5

    and then I end up with 28 degrees for the angle @ A. Maybe I'm looking at it the wrong way but beyond that I have no idea how to proceed with this question.

    Any help ?

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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by Monocerotis View Post
    So this is the triangle I'm dealing with.

    I have to find the measures of the angles x & y.

    First thing I tried to do was find the measure of the angle @ A.

    a/SinA = c/SinC
    66/SinA = 25/Sin10.5

    and then I end up with 28 degrees for the angle @ A. Maybe I'm looking at it the wrong way but beyond that I have no idea how to proceed with this question.

    Any help ?

    (angle C + angle A) - 180 = angle B
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  3. #3
    Newbie Monocerotis's Avatar
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    Quote Originally Posted by bigwave View Post
    (angle C + angle A) - 180 = angle B
    and....?
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  4. #4
    Super Member bigwave's Avatar
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    Quote Originally Posted by Monocerotis View Post
    and....?
    if you know angle B (or y) the angle x is 180 - y

    I redrew the triangle with a cad program this is what it looks with the angles distances are in feet


    is this any help?
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  5. #5
    Newbie Monocerotis's Avatar
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    The issue here is that I'm getting 28 degrees with the calculation 66/SinA = 25/Sin10.5

    the measurement for angle A is obviously not 28 degrees, so basically I'm ending up with an error and I can't understand why because so far as I understand my procedure is correct.

    I want to solve for angle A, then because I already have angle C I can solve for angle B.

    From angle B I could find the angle next to (x), thereby finding x becasue the sum of the two would be 180.

    makes sense now ?
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  6. #6
    Senior Member
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    Yes I get 28 as well.

    The only thing I can think of is the 10.5 angle is wrong or one of the lengths is wrong.

    Doesn't matter, just keep on going.

    So  180 - (28.76+10.5) = 140.74

    So  y =140.74

    Now use cosine law to find the right hand side then you can work out x.
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  7. #7
    Super Member
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    Quote Originally Posted by Monocerotis View Post
    So this is the triangle I'm dealing with.

    I have to find the measures of the angles x & y.

    First thing I tried to do was find the measure of the angle @ A.

    a/SinA = c/SinC
    66/SinA = 25/Sin10.5

    and then I end up with 28 degrees for the angle @ A. Maybe I'm looking at it the wrong way but beyond that I have no idea how to proceed with this question.

    Any help ?

    Your thinking is correct: the angle at A is 28.7574degrees.
    (see my attached image)
    But it is NOT the angle CAB, it is the angle EAB or angle CEB


     <br />
\angle CEB = \dfrac{66 \sin(10.5deg)}{25} = 28.7574 deg<br />

    \angle CEB = \angle EAB

    \angle CAB = 180 - 28.7574 = 151.2426 <br />

     <br />
y = \angle ABD = 180 - (10.5 + 151.2426) = 180 - 161.7426 = 18.2574<br />

     x = \angle ADC = 161.7426

    That is what is shown on the post by bigwave.
    Attached Thumbnails Attached Thumbnails Find measures of angles, massive error !-tripletri1.jpg  
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