7.1 Angles  (Page 6/29)

Since $\mathrm{45°}$ is half of $\mathrm{90°},$ we can start at the positive horizontal axis and measure clockwise half of a $\mathrm{90°}$ angle.

Because we can find coterminal angles by adding or subtracting a full rotation of $\mathrm{360°},$ we can find a positive coterminal angle here by adding $\mathrm{360°}.$

We can then show the angle on a circle, as in [link] .

When working in degrees, we found coterminal angles by adding or subtracting 360 degrees, a full rotation. Likewise, in radians, we can find coterminal angles by adding or subtracting full rotations of $2\pi$ radians:

The angle $\frac{11\pi }{4}$ is coterminal, but not less than $2\pi ,$ so we subtract another rotation.

The angle $\frac{3\pi }{4}$ is coterminal with $\frac{19\pi }{4},$ as shown in [link] .

1. Let’s begin by finding the circumference of Mercury’s orbit.
   $$\begin{array}{ccc}\hfill C& =& 2\pi r\hfill \\ & =& 2\pi (\text{36 million miles})\hfill \\ & \approx & \text{226 million miles}\hfill \end{array}$$

   Since Mercury completes 0.0114 of its total revolution in one Earth day, we can now find the distance traveled.

   $\left(0.0114\right)226\text{ million miles = 2}\text{.58 million miles}$
2. Now, we convert to radians.
   $\begin{array}{ccc}\hfill \text{radian}& =& \frac{\text{arclength}}{\text{radius}}\hfill \\ & =& \frac{2.\text{58 million miles}}{36\text{ million miles}}\hfill \\ & =& 0.0717\hfill \end{array}$