Question

Find all angles, osO<360, that satisfy the equation below, to thenearest 10th of a degree.cos(0) =-1/6

Answer 1

The cosine of the angle given is negative.

Cosine is negative in the second and third quadrant

`\begin{gathered} 0\leq x\leq360 \\ \cos (\theta)=-(1)/(6) \\ \theta=\cos ^(-1)((1)/(6)) \\ \theta=80.41^0 \\ \theta=80.4^0\text{ (to the nearest tenth)} \\ \end{gathered}`

In the second quadrant,

`\begin{gathered} 180-\theta \\ 180-80.4=99.6 \\ \theta=99.6^0 \end{gathered}`

In the third quadrant,

`\begin{gathered} 180+\theta \\ 180+80.4=260.4 \\ \theta=260.4^0 \end{gathered}`

Therefore, the values that satisfy the equation are 99.6 and 260.4 degrees.