Question

Hi, can you help me to solve this exercise, please!!!

Answer 1

Given:

`\begin{gathered} \cos (\theta)=-\frac{\sqrt[]{2}}{2} \\ \text{where;} \\ 0\le\theta\le\pi \end{gathered}`
`\begin{gathered} \cos (\theta)=-\frac{\sqrt[]{2}}{2} \\ \text{This shows the cosine of the angle has a negative value.} \\ Co\sin e\text{ is negative in the second and third quadrants.} \\ \text{Thus,} \\ \cos (\theta)=-\frac{\sqrt[]{2}}{2} \\ \cos \theta=\frac{\sqrt[]{2}}{2} \\ \cos \theta=(1.4142)/(2) \\ \cos \theta=0.7071 \\ \theta=\cos ^(-1)(0.7071) \\ \theta=45^0 \end{gathered}`

In the second quadrant,

`\begin{gathered} \theta=180-\theta \\ \theta=180-45 \\ \theta=135^0 \\ \\ \text{Taking this to radians,} \\ 2\pi\text{ radians=360 degr}ees \\ 2\pi=360^0 \\ x=135^0 \\ \text{Hence, the angle 135 degre}es\text{ in radians is,} \\ x=(2*\pi*135)/(360) \\ x=(270\pi)/(360)radians \\ x=(3\pi)/(4)\text{radians} \\ \\ \text{Hence, } \\ \theta=(3\pi)/(4)\text{radians} \end{gathered}`

We do not need the angle in the third quadrant since the range given is between 0 to 180 degrees and the third quadrant exceeds 180 degrees.

Therefore, the value of the angle in radians in simplified rationalized form is;

`\theta=(3\pi)/(4)\text{radians}`