Recall the formula for tangent is .  If we have enough information to solve for sine and cosine, then we have the values of the opposite and adjacent of the angle.

Consider the figure below.

We can form a triangle by dropping a line down from the point (-2,3) perpendicular to the 

x axis.  Now we have right triangle that has a leg that is 3 units high and a base that is 2 

units long.  Now we can use the Pythagorean Theorem to solve for the hypotenuse.

We can solve for sine and cosine now.  Use the fact that .  The opposite side of the angle in this case is the leg that is 3 units high.

And we will use the fact that .  The adjacent side to the angle is the base that is 2 units long.  We do have to note that we are using and not .

So even though our angle was obtuse, we can still use the same method.

We begin by thinking about what cosine means when working with right triangles.  Recall .  So if

Then the adjacent side, or the base of the triangle, is 1 and the hypotenuse is 2.  Now we can use the Pythagorean Theorem to solve for the missing side. 

Using the point (3,4), we can see that this forms a right triangle that has a base that is 3 units in length and an adjoining leg that is 4 units high.  We are able to find the hypotenuse of this triangle using the Pythagorean Theorem.  This will give us the distance of the point (3,4) to the origin. 

And so the hypotenuse of this triangle (the distance from our point we are working with to the origin), is 5 units long.  Recall that when using cosine for right triangles, cosine represents the following

So if we are considering the angle formed by the x-axis and our hypotenuse, the adjacent side would be the base of our triangle; 3 units. 

We see that the point on the terminal side is (5,6).  So we know that with this point a right triangle is formed with a base that is 5 units long, and a leg that is 6 units high.  We can use the Pythagorean Theorem to solve for the hypotenuse that is formed by this triangle and this will tell us the distance of the point from the origin. 

Let’s solve for sine first.  When working with right triangles recall that and we are considering the angle formed by the x-axis and the hypotenuse.  So the opposite side is the leg that is 6 units high.

We can solve for cosine if we recall that .  Our adjacent side would be the base that is 5 units long.

Using the equation,

  where 

=angular velocity, =linear velocity, and =radius of the circle.

In this case the radius is 5 (half of the diameter) and linear velocity is 20 m/s.  

.

Write the formula for angular velocity.

The frequency of the tire is 8 revolutions per second. The radius is not used.  

Substitute the frequency and solve.

Write the formula for average velocity.

The units of omega is radians per second.

Substitute the givens and solve for omega.

Recall that  .

Since the tire revolves 9.3 times/second it would seem that the tire would rotate

 or .

We use  to indicate that the tire is rolling 360 degrees or  radians each revolution (as it should).

Thus, 

 is your final answer.

Note that radians is JUST a different way of writing degrees. The higher numbers in the answers above are all measures around the actual linear speed of the tire, not the angular speed.

Angular speed is the same as linear speed, but instead of distance per unit time we use degrees or radians. Any object traveling has both linear and angular speed (though objects only have angular speed when they are rotating).

Since our tire completes 8 revolutions per second we multiply by  since a full rotation (360°) equals .