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Mathematics 14 Online
OpenStudy (anonymous):

Suppose a triangle has two sides of length 42 and 35, and that the angle between these two sides is 120. What is the length of the third side of the triangle?

OpenStudy (anonymous):

OpenStudy (anonymous):

help please

OpenStudy (anonymous):

can you help?

OpenStudy (anonymous):

hey m8

OpenStudy (anonymous):

hmm i will try

OpenStudy (anonymous):

Alright please

OpenStudy (arindameducationusc):

I am trying too...

OpenStudy (anonymous):

Thank you both.

OpenStudy (anonymous):

I am like stressing because of this test I cant pass

OpenStudy (anonymous):

ok so you use the law of cosine i think

OpenStudy (anonymous):

yea I did but I got lost after

OpenStudy (anonymous):

\[c^2=a^2+b^2\]

OpenStudy (anonymous):

oh i forgot the -2abcos(o)

OpenStudy (anonymous):

anyways \[c^2=32^2+35^2−2(32)(35)\cos(120)\]

OpenStudy (arindameducationusc):

cos A=b^2+c^2-c^2/2ab This is the cosine formula, Isn't it? @Icedragon50

OpenStudy (anonymous):

where did you get 32?

OpenStudy (anonymous):

\[c^2=1024+1225-2240(-\frac{ 1 }{ 2 })=3369\]

OpenStudy (anonymous):

c=\[√3369=58.04\]

OpenStudy (anonymous):

and there ya go!

OpenStudy (anonymous):

@arindameducationusc ya

OpenStudy (anonymous):

Thank you so much m8

OpenStudy (anonymous):

np :)

OpenStudy (anonymous):

g'night I'm tired xD

OpenStudy (anonymous):

Same its 1 lol

OpenStudy (anonymous):

bout to play league of legends

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