Question

given that $cos\theta=\frac{sqrt{50}}{10}$ and $\tan\theta$ is negative, determine $\theta$ and $\tan\theta$. enter the angle $\theta$ in degrees from the interval $0^{circ},360^{circ})$. write the exact answer. do not round.

Explanation:

Step1: Determine the quadrant

Since $\cos\theta=\frac{\sqrt{50}}{10}=\frac{\sqrt{2}}{2}>0$ and $\tan\theta < 0$, $\theta$ is in the fourth - quadrant.

Step2: Find the reference angle

We know that $\cos\theta=\frac{\sqrt{2}}{2}$, and the reference - angle $\theta_{r}$ for which $\cos\theta_{r}=\frac{\sqrt{2}}{2}$ is $\theta_{r} = 45^{\circ}$.

Step3: Calculate the angle $\theta$ in the fourth - quadrant

In the fourth - quadrant, $\theta=360^{\circ}-\theta_{r}$. So $\theta = 360^{\circ}-45^{\circ}=315^{\circ}$.

Step4: Calculate $\tan\theta$

We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Also, $\sin\theta=-\sin\theta_{r}=-\frac{\sqrt{2}}{2}$ (in the fourth - quadrant) and $\cos\theta=\frac{\sqrt{2}}{2}$. Then $\tan\theta=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=- 1$.

Answer:

$\theta = 315^{\circ}$, $\tan\theta=-1$