Question

Find cot0 if sin0 = -2/5 and 0 terminates in the third quadrent

Answer 1

`cot\: \theta=\: \frac{\sqrt[]{21}}{2}`

Explanation:

We have the following information.

`\begin{gathered} \sin \theta=-(2)/(5)=\frac{\text{ opposite}}{\text{ hypotenuse}} \\ \text{opposite = 2} \\ \text{ hypotenuse = 5} \end{gathered}`

We can calculate the adjacent leg by means of Pythagoras' theorem, just like this:

`\begin{gathered} h^2=a^2+b^2 \\ 5^2=2^2+b^2 \\ b^2=25-4 \\ b=\sqrt[]{21} \\ \text{therefore} \\ \cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}}=-\frac{\sqrt[]{21}}{5} \\ \text{replacing} \\ \cot \theta=(\cos\theta)/(\sin\theta)=\frac{-\frac{\sqrt[]{21}}{5}}{-(2)/(5)}=\frac{\sqrt[]{21}}{2} \\ \cot \theta=\frac{\sqrt[]{21}}{2} \\ \end{gathered}`

We determine that the angle is in the third quadrant

`\begin{gathered} cot\: \theta=\: (√(21))/(2) \\ \theta=\cot ^(-1)(\frac{\sqrt[]{21}}{2}) \\ \theta=203.58\text{\degree} \\ \text{ we check:} \\ \sin 203.58=-0.40002\cong-(2)/(5) \\ \cos 203.58=-0.91650\cong-\frac{\sqrt[]{21}}{5} \end{gathered}`