cot y, if csc y=-10/3 and cos y<0 Evaluate w/o using a calculator. No decimal answers and use pi radians if necessary.

help!!

help!

y=((180 (2 π ℕ_3 +((π arcsin(((3)/(10))))/(180))+π))/(π))

I don't understand that equation

u searching for y , ?

\[\csc(y)=-\frac{10}{3}\implies \sin(x)=-\frac{3}{10}\]

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we get by pythagoras third side is \[\sqrt{10^2-3^2}=\sqrt{100-9}=\sqrt{91}\] so \[\cos(y)=-\frac{\sqrt{91}}{10}\]

|dw:1338142851721:dw| well the y is more like this

\[\cot(y)=\frac{\sqrt{91}}{3}\]

doesn't matter what the y looks like. you want \[\cot(y)\] or \[\cot(\zeta)\] or whatever

\[\cot(\gamma)\]i bet

\[\cot \gamma\]

yes!

@jeessicaaahh name of variable is not an issue here. you are not looking for \(\gamma\) (it is the greek letter gamma) but you are looking for \(\cot(\gamma)\)

yes, yes i am.

you need to find all three sides of the right triangle, and then you can find it by looking

by looking .. ?

do i use the drawing you gave me to help?

yes find the third side

cosecant you can think of as "hypotenuse over opposite" so label the triangle that way, and then find the third side

close \[\cos(\gamma)=-\frac{\sqrt{91}}{10}\] you know it is negative because you are told

i mean you are told that it is negative

yes, so my answer is 3 ?

cotangent is adjacent over opposite, and that is what you are looking for

adjacent is \(\sqrt{91}\) and opposite is 3

got it?

i got 3 is that correct ?

no it is not 3, it is the ratio \(\frac{\sqrt{91}}{3}\)

adjacent over opposite

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