1. Law of Cosines

2. Let's consider types of triangles with the three pieces of information shown below. SAS You may have a side, an angle, and then another side AAA You may have all three angles. SSS You may have all three sides This case doesn't determine a triangle because similar triangles have the same angles and shape but "blown up" or "shrunk down" We can't use the Law of Sines on these because we don't have an angle and a side opposite it. We need another method for SAS and SSS triangles. AAA

3. Let's place a triangle on the rectangular coordinate system.  a b c (b, 0) What is the coordinate here? Drop down a perpendicular line from this vertex and use right triangle trig to find it. (x, y) (a cos , a sin  ) Now we'll use the distance formula to find c (use the 2 points shown on graph) square both sides and FOIL factor out a 2 This = 1 y x rearrange termsThis is the Law of Cosines

4. We could do the same thing if gamma was obtuse and we could repeat this process for each of the other sides. We end up with the following: LAW OF COSINES If you say it in words, you don't need to memorize 3 formulas: One side squared equals the sum of each of the other sides squared minus two times the product of those other sides times the cosine of the angle between those sides.

5. Since the Law of Cosines is more involved than the Law of Sines, when you see a triangle to solve you first look to see if you have an angle (or can find one) and a side opposite it. You can do this for ASA, AAS and SSA. In these cases you'd solve using the Law of Sines. However, if the 3 pieces of info you know don't include an angle and side opposite it, you must use the Law of Cosines. These would be for SAS and SSS (remember you can't solve for AAA).

6. Solve a triangle where b = 1, c = 3 and  = 80° Draw a picture. 80   a 1 3 Do we know an angle and side opposite it? No so we must use Law of Cosines. Hint: we will be solving for the side opposite the angle we know. This is SAS times the cosine of the angle between those sides One side squared sum of each of the other sides squared minus 2 times the product of those other sides Now punch buttons on your calculator to find a. It will be square root of right hand side. a = 2.99 CAUTION: Don't forget order of operations: powers then multiplication BEFORE addition and subtraction

7. We'll label side a with the value we found. We now have all of the sides but how can we find an angle? 80   2.99 1 3 Hint: We have an angle and a side opposite it. 80.8  is easy to find since the sum of the angles is a triangle is 180° 19.2 NOTE: These answers are correct to 2 decimal places for sides and 1 for angles. They may differ with book slightly due to rounding. Keep the answer for  in your calculator and use that for better accuracy.

8. Solve a triangle where a = 5, b = 8 and c = 9 Draw a picture.   5 8 9 Do we know an angle and side opposite it? No, so we must use Law of Cosines. Let's use largest side to find largest angle first. This is SSS times the cosine of the angle between those sides One side squared sum of each of the other sides squared minus 2 times the product of those other sides CAUTION: Don't forget order of operations: powers then multiplication BEFORE addition and subtraction  84.3

9. How can we find one of the remaining angles?   5 8 9 Do we know an angle and side opposite it?  84.3 62.2 33.5 Yes, so use Law of Sines.

10. Length of Chord A chord is a line segment inside a circle with endpoints on the circle. If we wanted to find the length a of the chord shown on the left, can you think of something you just learned to help you find this length? The Law of Cosines! Equation for Length of Chord

11. So to find the length of a chord, we’d need the radius of the circle r and the central angle  which is the angle formed with the center of the circle as the vertex and rays containing the endpoints of the chord.

12. Find the length of the side of a regular hexagon inscribed in circle of radius 5. 5  a Let’s get a picture of this. We’ve got r but how do we find  ? We could also have arrived at this seeing that we have an equilateral triangle.