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    Right.
    I've got a bit of a problem here. I've been given a problem, and I have to work out the rest sides and angle in a triangle.
    However, the question says that there are 2 possible triangles are possible.
    This is the information I've been given:
    Angle A = 28
    Side a = 10.6 cm
    Side c = 12.8 cm

    Using the sin rule I've worked out
    Angle B = 117.465
    Angle C = 34.535
    Side b = 20.03

    I cannot work out how there could be another triangle from this information. So any help at all would be welcome.

    Matchless
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    When you try to find the angle C, you end up with something like:

    c =  \arcsin  0.xxx

    34, you said, is one answer. But there is also another answer, which is 180 minus, since this is sine positive.
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    (Original post by 2710)
    When you try to find the angle C, you end up with something like:

    c =  \arcsin  0.xxx

    34, you said, is one answer. But there is also another answer, which is 180 minus, since this is sine positive.
    But we have to find another real triangle, and draw it as well.
    I thought that we wouldn't be able to do it like that. Unless I'm more stupid at maths than I thought.
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    One of the sides can be in 2 possible positions. The length of the bottom side (in the diagram) will be different for the 2 triangles.
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    To complement what others have said, the reason why there are 2 values for sin^-1(whatever the value is) is because, if you look at a sine wave, y=that value intersects twice with y=sin(x)
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    I wish I studied more now.
    Thanks people.
 
 
 
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Updated: January 14, 2010
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