Question

angle t has a measure between 0° and 360° and is coterminal with a -710° angle. what is the measure of angle t? 5° 10° 15° 20°

Explanation:

Step1: Recall coterminal angle formula

To find a coterminal angle with a given angle \(\theta\) (in degrees), we add or subtract multiples of \(360^\circ\) until we get an angle between \(0^\circ\) and \(360^\circ\). For a negative angle \(-710^\circ\), we will add multiples of \(360^\circ\) to it. Let \(n\) be the number of times we add \(360^\circ\). We need to find \(n\) such that \(-710^\circ + n\times360^\circ\) is between \(0^\circ\) and \(360^\circ\).

Step2: Find the appropriate multiple of \(360^\circ\)

Let's solve for \(n\) in the inequality \(0^\circ\leq - 710^\circ + n\times360^\circ\leq360^\circ\).
First, add \(710^\circ\) to all parts: \(710^\circ\leq n\times360^\circ\leq710^\circ + 360^\circ=1070^\circ\).
Now, divide all parts by \(360^\circ\): \(\frac{710}{360}\leq n\leq\frac{1070}{360}\).
\(\frac{710}{360}\approx1.97\) and \(\frac{1070}{360}\approx2.97\). Since \(n\) must be an integer, \(n = 2\).

Step3: Calculate the coterminal angle

Substitute \(n = 2\) into the formula: \(-710^\circ+2\times360^\circ=-710^\circ + 720^\circ = 10^\circ\). Wait, no, wait, let's check again. Wait, \(2\times360 = 720\), \(-710+720 = 10\)? Wait, but let's check with \(n = 3\): \(3\times360=1080\), \(-710 + 1080=370\), which is more than \(360\). Wait, maybe I made a mistake in \(n\). Let's try \(n = 2\): \(-710+720 = 10\), but let's check the number of full rotations. The absolute value of \(-710\) is \(710\). Divide \(710\) by \(360\): \(710\div360 = 1\) with a remainder of \(350\)? No, \(360\times1 = 360\), \(710 - 360=350\), so \(-710=-360\times2 + 10\)? Wait, \(-360\times2=-720\), \(-720 + 10=-710\). So the coterminal angle is \(10^\circ\)? Wait, but let's recalculate:

Wait, to find the coterminal angle of a negative angle, we can keep adding \(360^\circ\) until we get a positive angle between \(0\) and \(360\).

Start with \(-710^\circ\). Add \(360^\circ\): \(-710 + 360=-350^\circ\) (still negative). Add another \(360^\circ\): \(-350+360 = 10^\circ\). Ah, there we go. So after adding \(360^\circ\) twice, we get \(10^\circ\), which is between \(0^\circ\) and \(360^\circ\).

Wait, but let's check the options. The options are \(5^\circ\), \(10^\circ\), \(15^\circ\), \(20^\circ\). So the answer should be \(10^\circ\). Wait, but let me check again. Wait, \(-710 + 360\times2=-710 + 720 = 10\). Yes, that's correct.

Answer:

\(10^\circ\) (corresponding to the option: 10°)