# Deriving the Law of Sines

## Behold.

A circle with an inscribed triangle.

## Deriving the Law of Sines

Let's work on that diagram.
• If you're a web-browsing wizard, you might open this activity in two side-by-side views so you can keep the diagram in one view and the instructions/Q&A in the other.

## 1. Inscribed Angle Theorem

The Inscribed Angle Theorem says that an inscribed angle has half the measure of a central angle intercepting the same arc. For example, angle ABC has half the measure of angle ADC.
Something funny happens because of the Inscribed Angle Theorem. Let's check it out:
• Use the Angle tool to measure angle ABC.
• As you move vertex B around the circle, the measure of angle ABC doesn't change until it passes A or C. At that moment, the arc intercepted by angle ABC flips from one side of AC to the other.

## 2. Angle bisector at B

1. Construct the bisector of angle B like this: behind the Perpendicular Bisector tool, find the Angle Bisector tool; then click on segments a and c.
2. Oh look, GeoGebra gave you the bisector you wanted and...a line perpendicular to it? Why? Well, the segments you clicked on are parts of lines, and those lines meet in an X-shape that has four angles. To bisect all four of those angles, we needed four two bisectors. Wait...if you bisect two adjacent supplementary angles, are the bisectors always perpendicular? Hm.
3. OK, now move B around. The angle bisectors seem to be really interested in two special points on the circle. I mean, there are two points we haven't named yet, that the bisectors of angle B keep going through. Move B faster...maybe you can trick them.
4. No? Did you try moving B past A or C? What is the deal with those two points?

## Q/A #1

Strange things are happening. Why do we need only two lines to bisect four angles? Are the bisectors of a linear pair always perpendicular? Do you notice anything interesting about the two mystery points? Respond to any or all of these questions. If this is a GeoGebra Classroom activity, your teacher will be able to react to your answers. If this is NOT a GeoGebra Classroom activity, you will have given some electrons a little bit of joy before sending them into the ether.

## 3. Perpendicular bisector of the side opposite B

Construct the perpendicular bisector of segment b. The Perpendicular Bisector tool is behind the Angle Bisector tool, right where you put it. Now when you move B around (remember to switch to the Move tool), you should see a black-framed right triangle whose hypotenuse is a diameter of the circle. This is exactly what it looks like, but my proof of that fact is another detour. Here is the future home of a link to it.

## The three steps above are background information.

Once you're comfortable with those ideas, reset the construction and move B so that AB (a.k.a c) is a diameter.

## Q/A #2

What's the measure of angle C right now? If you're not sure, think about the Inscribed Angle Theorem again.

## Q/A #3

Now AB is a diameter of the circle and there's a right angle at C. That means ABC IS A RIGHT TRIANGLE oops, fixed my caps lock. How's your right triangle trig? I mean, what's ?

## Q/A #4

What's a pirate's favorite letter?

## Q/A #5

On the advice of Q/A #4, solve for c in the equation from Q/A #3.

## Q/A #6

As you move point B around, does side b change?

## Q/A #7

As you move point B around, does the measure of change?

The two main points of interest from everything we've done so far:
1. When B is positioned just right, is the diameter of the circle; and
2. When you move B to other places, neither b nor sin(B) change value.
Therefore, the value of is always the diameter of the circle. Nothing here makes B any more special than A or C. So, and are also always the diameter of the circle. Since the three expressions , , and are three different ways to express the same quantity, they're all equal: . At last (drum roll, please): Equal, nonzero fractions have equal reciprocals, so , which we call the Law of Sines.