Question

Given that sino =V48and cotê is negative, determine 0 and coté. Enter the angle O in degrees from the interval [0°, 360). Write the exact answer. Do not round.

Answer 1

In this problem

we have that

sin(theta) is positive and cos(theta) is negative

That means

the angle theta lies on the II quadrant

Remember that

`\cot (\theta)=(\cos(\theta))/(\sin(\theta))`

Find out the value of cos(theta)

`\sin ^2(\theta)+\cos ^2(\theta)=1`

substitute the given value

`(\frac{\sqrt[]{48}}{8})^2+\cos ^2(\theta)=1`
`\cos ^2(\theta)=1-(48)/(64)`
`\begin{gathered} \cos ^2(\theta)=(16)/(64) \\ \cos ^{}(\theta)=-(4)/(8) \end{gathered}`

Find out the value of cot(theta)

substitute given values

`\cot (\theta)=-\frac{4}{\sqrt[\square]{48}}`

simplify

`\cot (\theta)=-\frac{4}{\sqrt[\square]{48}}\cdot\frac{\sqrt[]{48}}{\sqrt[]{48}}=-\frac{4\sqrt[]{48}}{48}=-\frac{\sqrt[]{48}}{12}=-\frac{4\sqrt[]{3}}{12}=-\frac{\sqrt[]{3}}{3}`

Find out the angle theta

using a calculator

angle in II quadrant

theta=120 degrees

Convert to radians ---->