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tan 40-9.8*20^2/(92*16^2*cos^2*40)

 Feb 24, 2016
 #1
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Cos of what?

I'm gonna assume it is 40 with the tan

\(tan{40}-9.8(\frac{20^2}{92*16^2*cos^2{40}*40})\)

tan(40)-9.8(20^2/92*16^2*cos(40)*cos(40) *40) = -194075.942

-194075.942

 Feb 24, 2016
 #2
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I would have assummed they meant (cos(40))^2   read as cos squared 40 or cos 40 all squared

 

\(tan 40-9.8*20^2/(92*16^2*(cos(40))^2)\)

 

That would give 0.555471

 

But it is a guess at what the cos^2*40 meant

 Feb 24, 2016
edited by Guest  Feb 25, 2016

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