find the values of angles X, Y, and Z.
x=91; y= 51;z=31
x=89; y=91; z=0
x=60; y=120; z=31
please HELP!!! math 7 B unit 1 geometry test!!!! anyone from Connections Academy 7th grade please help me

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  1. First off, each side must be less than the sum of the other two. None of your "triangles" can exist. Are they sides of some other polygon?

    However, given three sides x,y,z of a triangle XYZ, you can find the angles in several ways. One of the easiest is to use the law of cosines first. So, let's say you have sides 51,31,71. Then

    x^2 = y^2 + z^2 - 2yz cosX
    cosX = (y^2+z^2-x^2)/(2yz)
    = (51^2 + 31^2 - 71^2)/(2*51*31)
    = -29/62
    so, X is 117.9°

    Now, you do Y and Z the same way, or use the law of sines to find Y:

    sinY/y = sinX/x
    sinY/51 = 0.8838/71
    sinY = 0.6348
    Y = 39.4°

    Now, Z = 180-X-Y = 22.7°

    So, fix up your data, and then you can find the angles you need.

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  2. you confused me sooooo much

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  3. Ah i see thank you

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  4. That is so wrong!!! i am so dead now

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  5. um, this didnt help.

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