Asked by georgia :)

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
Answered by Steve
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.
Answered by Hello
you confused me sooooo much
Answered by Jane
Ah i see thank you
Answered by #allstar
That is so wrong!!! i am so dead now
Answered by do kyungsoo
um, this didnt help.
Answered by Jack
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
Answered by Kitty
Which ever angles in the answers that add up to 180 are the answers.
Answered by Baby yoda
It’s 20 (A)
Answered by Anonymous
thank you
Answered by Xzavier dentrell king
BRO lets goooooooo
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