Question

question
for the rotation $-918^{circ}$, find the coterminal angle from $0^{circ} \leq \theta < 360^{circ}$, the quadrant, and the reference angle.

answer
attempt 1 out of 2

the coterminal angle is \\(\square^{\circ}\\), which lies in quadrant \\(\boldsymbol{\downarrow}\\), with a reference angle of \\(\square^{\circ}\\).

Explanation:

Step1: Find coterminal angle

To find a coterminal angle of \(-918^{\circ}\) in the range \(0^{\circ}\leq\theta < 360^{\circ}\), we add multiples of \(360^{\circ}\).
Let \(n\) be an integer. We want to solve \(-918 + 360n\geq0\) and \(-918+360n < 360\).
First, solve \(-918 + 360n\geq0\): \(360n\geq918\), \(n\geq\frac{918}{360}=2.55\).
Then solve \(-918 + 360n<360\): \(360n<360 + 918=1278\), \(n<\frac{1278}{360}=3.55\).
So \(n = 3\).
\(\theta=-918^{\circ}+3\times360^{\circ}=-918^{\circ}+1080^{\circ}=162^{\circ}\)

Step2: Determine the quadrant

Since \(90^{\circ}<162^{\circ}<180^{\circ}\), the angle \(162^{\circ}\) lies in Quadrant II.

Step3: Calculate the reference angle

For an angle \(\theta\) in Quadrant II, the reference angle \(\alpha\) is given by \(\alpha = 180^{\circ}-\theta\).
So \(\alpha=180^{\circ}-162^{\circ}=18^{\circ}\)

Answer:

The coterminal angle is \(162^{\circ}\), which lies in Quadrant II, with a reference angle of \(18^{\circ}\).