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

$\cot (\theta)=\frac{\cos(\theta)}{\sin(\theta)}$

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$$\cos ^2(\theta)=1-\frac{48}{64}$$\begin{gathered} \cos ^2(\theta)=\frac{16}{64} \\ \cos ^{}(\theta)=-\frac{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 ---->