Question

finding coterminal angles
this is the only question in this section.
question
find an angle \\( \theta \\) coterminal to \\( 550^\circ \\), where \\( 0^\circ \leq \theta < 360^\circ \\).

Explanation:

Step1: Recall coterminal angle formula

To find a coterminal angle with a given angle \( \alpha \) (in degrees) within the range \( 0^\circ \leq \theta < 360^\circ \), we subtract or add multiples of \( 360^\circ \) until we get an angle in that range. The formula is \( \theta=\alpha - n\times360^\circ \) (where \( n \) is a positive integer chosen such that \( \theta \) is in the desired range).

For \( \alpha = 550^\circ \), we need to find \( n \) such that \( 0^\circ \leq 550^\circ - n\times360^\circ< 360^\circ \).

Step2: Determine the value of \( n \)

Let's try \( n = 1 \): \( 550^\circ- 1\times360^\circ=550 - 360=190^\circ \). Wait, no, wait, \( 550-360 = 190? \) Wait, no, \( 550-360 = 190? \) Wait, no, \( 550-360=190? \) Wait, no, \( 550 - 360=190 \)? Wait, no, \( 550-360 = 190 \)? Wait, no, let's calculate again. \( 550-360 = 190 \)? Wait, no, \( 360\times1 = 360 \), \( 550 - 360=190 \)? Wait, no, \( 550-360 = 190 \)? Wait, no, that can't be. Wait, no, \( 360\times2=720 \), which is more than 550, so \( n = 1 \) gives \( 550 - 360=190 \)? Wait, no, \( 550-360 = 190 \)? Wait, no, \( 550-360 = 190 \)? Wait, no, I made a mistake. Wait, \( 550-360 = 190 \)? Wait, no, \( 360 + 190=550 \), yes. Wait, but \( 190 \) is between \( 0 \) and \( 360 \)? Wait, \( 0\leq190 < 360 \), yes. Wait, but wait, is that correct? Wait, no, wait, \( 550^\circ \) is more than \( 360^\circ \), so to find the coterminal angle, we subtract \( 360^\circ \) once: \( 550 - 360=190^\circ \). Wait, but let's check: \( 190^\circ+360^\circ = 550^\circ \), so they are coterminal. Wait, but wait, is \( 190^\circ \) in the range \( 0^\circ\leq\theta < 360^\circ \)? Yes, because \( 0\leq190 < 360 \). Wait, but wait, maybe I messed up. Wait, no, let's do it properly. The general formula for coterminal angles (in degrees) is \( \theta=\alpha + k\times360^\circ \), where \( k \) is any integer (positive or negative). We need \( 0^\circ\leq\theta < 360^\circ \). So for \( \alpha = 550^\circ \), we solve \( 0\leq550 + k\times360< 360 \). Let's solve for \( k \). Subtract 550: \( - 550\leq k\times360<360 - 550=-190 \). Divide by 360: \( \frac{-550}{360}\leq k<\frac{-190}{360} \). \( \frac{-550}{360}\approx - 1.527 \), \( \frac{-190}{360}\approx - 0.527 \). So \( k=-1 \). Then \( \theta=550+(-1)\times360 = 550 - 360 = 190^\circ \).

Answer:

\( 190^\circ \)