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How would I divide one equation by another?
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\[(250*\sin(38)-\sin(\theta)m)/(250*\cos(38)+mcos(\theta)-1000)\] for example. I never really got my head around this,
would it be like 250∗sin(38)/250∗cos(38) - tan(theta) - 1/1000
Oh gosh I wish I could help. \[\frac{ (250*\sin(38)-\sin(\theta)m)}{ (250*\cos(38)+mcos(\theta)-1000) }\] @jim_thompson5910
;)
the real question is this, 0 = 250*cos(38)+fcos(a)-100 -1 0 = 250*sin(38)-sin(theta)f -2 I want to find theta and f. so I divide the equations right 2/1
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were theta + a are the same variables. soz about that
Since they both equal zero, I think I'd try substitution / transitive property. The expressions on the right must equal each other. If a=c and b=c then a=b.
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