Question

Find cot θ if csc θ = square root of seventeen divided by four and tan θ > 0

Answer 1

\(\bf csc(\theta) = \cfrac{\sqrt{17}}{4}\quad ?\)

Answer 2

im sorry what?

Answer 3

yes it has that then after that it says "and tan θ > 0"

Answer 4

ok

Answer 5

you know how to do it?

Answer 6

well notice that \(\bf csc(\theta) = \cfrac{\sqrt{17}}{4} \implies \cfrac{\textit{hypotenuse}}{\textit{opposite side}}
\implies \cfrac{c}{b}\\\quad \\
\textit{now get the "a" or adjacent side, keeping in mind that }\\
c^2 = a^2 + b^2 \implies \sqrt{c^2-b^2} = a\)

Answer 7

\(\bf csc(\theta) = \cfrac{\sqrt{17}}{4} \implies \cfrac{\textit{hypotenuse}}{\textit{opposite side}}
\implies \cfrac{c}{b}\\\quad \\
c = \sqrt{17} \qquad \qquad b = 4\)

Answer 8

ok I see that what do I do to start solving it

Answer 9

thank you btw

Answer 10

well, you see, you're asked for the cotangent function
the cotangent is \(\bf cot(\theta) = \cfrac{adjacent}{opposite} \implies \cfrac{a}{b}\)

so, we would need the "a" component, that is, the adjacent side, to get the cotangent function

Answer 11

we know what "c" is, we know what "b" is, so, using the pythagorean theorem \(\bf c^2 = a^2 + b^2 \implies \sqrt{c^2-b^2} = a\)

Answer 12

so its 1?

Answer 13

hmmm, well, yes, is unit 1
now, we also know that the tangent of that angle is > 0, that is, is POSITIVE
the tangent is only positive in 1st Quadrant and 3rd Quadrant

now, also notice, the hypotenuse is always positive
so if say \(\bf csc(\theta) = \cfrac{\sqrt{17}}{4} \implies \cfrac{\sqrt{17}}{+4}\)

that means that the side "b" is positive, side "b" is the y-coordinate, so our "y" is positive
now, where between 1st and 3rd Quadrant is the "y" positive? well in the 1st one

so our angle is in the 1st Quadrant, that means the "a" side, that is our x-coordinate, is also positive on that quadrant, so our 1 from the pythagorean theorem, is +1

so, what do you think is our cotangent?

Answer 14

ok im a little lost where ami looking to find the answer?

Answer 15

hmmm... .well, check your trig identities for cotangent

Answer 16

is it 1/4 the answer?

Answer 17

\(\bf csc(\theta) = \cfrac{\sqrt{17}}{4} \implies \cfrac{\textit{hypotenuse}}{\textit{opposite side}}
\implies \cfrac{c}{b}\\\quad \\
c = \sqrt{17} \qquad \qquad b = \color{red}{4}\qquad \qquad a = \color{blue}{1}\\\quad \\
cot(\theta) = \cfrac{adjacent}{opposite} \implies \cfrac{a}{b} \implies \cfrac{\color{blue}{1}}{\color{red}{4}}\)

Answer 18

yeap

Answer 19

sweet thank you :)

Answer 20

yw