Question

I really need help solving this practice from my prep guide in trigonometry

Answer 1

Given: Different angles in degrees and in terms of pi. The different angles are:

`\begin{gathered} a)714^0 \\ b)(23\pi)/(5) \\ c)120^0 \\ d)(31\pi)/(6) \end{gathered}`

To Determine: The equivalence of the given angles

The equivalent of degree and pi is given as

`\begin{gathered} 2\pi=360^0 \\ \pi=(360^0)/(2) \\ \pi=180^0 \\ 360^0=2\pi \\ 1^0=(2\pi)/(360^0) \\ 1^0=(1)/(180)\pi \end{gathered}`
`\begin{gathered} a)714^0 \\ 1^0=(1)/(180)\pi \\ 714^0=(714^0)/(180^0)\pi \\ 714^0=3(29)/(30)\pi \\ 714^0=\frac{119\pi^{}}{30} \end{gathered}`
`\begin{gathered} b)(23\pi)/(5) \\ 1\pi=180^0 \\ (23\pi)/(5)=(23)/(5)*180^0 \\ (23\pi)/(5)=828^0 \end{gathered}`
`\begin{gathered} c)120^0 \\ 1^0=(\pi)/(180) \\ 120^0=120*(\pi)/(180) \\ 120^0=(2\pi)/(3) \end{gathered}`
`\begin{gathered} d)(31\pi)/(6) \\ 1\pi=180^0 \\ (31\pi)/(6)=(31)/(6)*180^0 \\ (31\pi)/(6)=930^0 \end{gathered}`

ALTERNATIVELY

A revolution is 360 degree

`\begin{gathered} a)714^0 \\ \text{Multiples of 360 degre}e \\ 2*360^0=720^0 \\ \text{equivalent of 714 degre}e\text{ would be} \\ 720^0-714^0=6^0 \end{gathered}`

`\begin{gathered} a)714^0=(119\pi)/(30) \\ b)(23\pi)/(5)=828^0 \\ c)120^0=(2\pi)/(3) \\ d)(31\pi)/(6)=930^0 \end{gathered}`