Question

question
for the rotation $-1057^circ$, 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\\)°, which lies in quadrant \\(\square\\), with a reference angle of \\(\square\\)°.

Explanation:

Step1: Find coterminal angle

To find a coterminal angle with \(-1057^\circ\) in the range \(0^\circ\leq\theta < 360^\circ\), we add multiples of \(360^\circ\) until we get into that range. Let's find how many times \(360^\circ\) we need to add.

First, divide \(1057\) by \(360\): \(1057\div360\approx2.936\). So we need to add \(3\) times \(360^\circ\) (since adding \(2\) times would not be enough).

Calculate \( - 1057+3\times360=-1057 + 1080 = 23^\circ\). Wait, no, wait: Wait, \(3\times360 = 1080\), \(-1057+1080 = 23\)? Wait, no, let's check again. Wait, maybe I miscalculated the number of times. Let's do it step by step.

We know that coterminal angles differ by \(360^\circ n\), where \(n\) is an integer. We want \(\theta=- 1057+360n\) such that \(0\leq\theta<360\).

Let's solve for \(n\):

\(0\leq - 1057 + 360n<360\)

Add \(1057\) to all parts:

\(1057\leq360n<1057 + 360=1417\)

Now divide by \(360\):

\(\frac{1057}{360}\leq n<\frac{1417}{360}\)

\(\frac{1057}{360}\approx2.936\) and \(\frac{1417}{360}\approx3.936\). So \(n = 3\)

Then \(\theta=-1057+3\times360=-1057 + 1080=23^\circ\)? Wait, no, that can't be right. Wait, wait, maybe I made a mistake in the sign. Wait, the angle is negative, so we are rotating clockwise. Let's check again.

Wait, let's compute how many full rotations (360 degrees) are in 1057 degrees. Let's do \(1057\div360 = 2\) with a remainder. \(2\times360 = 720\), \(1057 - 720=337\). Wait, no, if the angle is \(-1057^\circ\), to make it positive, we can add \(360^\circ\) repeatedly until we get a positive angle less than \(360^\circ\).

Let's add \(360^\circ\) once: \(-1057 + 360=-697^\circ\)

Add \(360^\circ\) again: \(-697+360 = - 337^\circ\)

Add \(360^\circ\) again: \(-337 + 360=23^\circ\). Oh, right, so the coterminal angle is \(23^\circ\).

Step2: Determine the quadrant

The angle \(23^\circ\) is between \(0^\circ\) and \(90^\circ\), so it lies in Quadrant I.

Step3: Find the reference angle

For an angle in Quadrant I, the reference angle is equal to the angle itself. So the reference angle for \(23^\circ\) is \(23^\circ\).

Wait, but wait, did I make a mistake in calculating the coterminal angle? Let's check again. Let's see, \(-1057^\circ\). Let's find how many times 360 goes into 1057. \(360\times2 = 720\), \(1057 - 720 = 337\). Wait, no, if we have a negative angle, adding 360 each time. Let's do:

\(-1057+360=-697\)

\(-697 + 360=-337\)

\(-337+360 = 23\). Yes, that's correct. So the coterminal angle is \(23^\circ\), Quadrant I, reference angle \(23^\circ\).

Wait, but let's verify. Let's check the number of rotations. The angle \(-1057^\circ\) is a clockwise rotation. A full clockwise rotation is \(360^\circ\), so two full clockwise rotations is \(720^\circ\), three full clockwise rotations is \(1080^\circ\). So starting from \(0^\circ\), rotating clockwise \(1057^\circ\) is the same as rotating counterclockwise \(1080 - 1057=23^\circ\). Yes, that makes sense. So the coterminal angle is \(23^\circ\), which is in Quadrant I, reference angle \(23^\circ\).

Answer:

The coterminal angle is \(\boldsymbol{23}^\circ\), which lies in Quadrant \(\boldsymbol{I}\), with a reference angle of \(\boldsymbol{23}^\circ\).