In Triangle XYZ angle x = 42, angle Y = 58 and z = 10. what is the measure of side x

i am going to fail this math takehome test.

Note that, since X, Y, and Z are the three angles of a triangle, then:
X + Y + Z = 180° ==> Z = 180° - X - Y.

Thus, Z = 180° - 42° - 58° = 80°.

Then, using the Law of Sines:
sin(X)/x = sin(Z)/z
==> sin(42°)/x = sin(80°)/10
==> x = 10sin(42°)/sin(80°) ≈ 6.8.

I hope this helps!

Use cosine rule to find y
y^2 = x^2 + z^2 - 2xz cos y
=42^2 + 10^2 - 2*10*42 cos 58
= 1764 +100 - 840 cos 58
= 1864 -840 cos 58
= 1418.86
y = 37.67

Now sine rule to find X
sin X / 42 = sin 58 / 37.67

sin X = 42 sin 58 / 37.67

X = sin^-1 [42 sin 58 / 37.67 ] = 71* 1'

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I think you need to measure side X...