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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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10 answers
  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
    so,
    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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  6. you take the left triangle, and add the angles you know, then subtract from 180 to determine x.

    you take the right side triangle (not a right triangle) and you take what you know, 38 degrees, add in 51 degrees add them together and subtract from 180,

    x=89, y=91 and z=51

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  7. Which ever angles in the answers that add up to 180 are the answers.

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  8. It’s 20 (A)

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  9. thank you

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  10. BRO lets goooooooo

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