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Mathematics 18 Online
OpenStudy (anonymous):

compare the measures.

OpenStudy (anonymous):

OpenStudy (anonymous):

Whats different?

OpenStudy (anonymous):

the angle. @Andrewthehelper

OpenStudy (anonymous):

Try using cosine law for both triangles: c^2 = a^2 + b^2 - 2abCos(C)

OpenStudy (anonymous):

for the 42 degree triangle SU is 7.2 and for the 43 degree VX is 7.12 ?

OpenStudy (anonymous):

@DoritosHero

OpenStudy (anonymous):

I got for the first one: c^2 = 4^2 + 5^2 - 2*4*5*Cos(42) SU = c = 3.35

OpenStudy (anonymous):

so it would be the same equation but cos(43) rather than cos(42) right?

OpenStudy (anonymous):

Yep!

OpenStudy (anonymous):

okay. one sec let me do the math then.

OpenStudy (anonymous):

4.3 is what i got but that doesnt seem right.

OpenStudy (anonymous):

wait. 3.42?

OpenStudy (anonymous):

\[(VX)^{2} = (VW)^{2} + (WX)^{2} - 2(VW)(WX)\cos (VX)\] \[(VX)^{2} = (5)^{2} + (4)^{2} - 2*5*4*\cos(43)\] \[(VX)^{2} = 11.75\] VX = 3.427

OpenStudy (anonymous):

Yes that's right

OpenStudy (anonymous):

omg i actually did it. lol

OpenStudy (anonymous):

Yes you did! Good job :)

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