Section 6.1 Law of Sines

General Comments A B C a b c We learned to solve right triangles in chapter 4. We will start this chapter by learning to solve oblique triangles (non-right triangles). Please note that angles are Capital letters and the side opposite is the same letter in lower case. C Make sure students understand the opposite side and the 2 adjacent sides (vs. adjacent and hypotenuse). a b B A c

What we already know   A B C a b c

Let’s look at an oblique triangle If we think of h as being opposite to both A and B, then A B C a b c h Let’s solve both for h. Let’s drop an altitude and call it h. This means…

A B C a b c If I were to drop an altitude to side a, I could come up with Putting it all together gives us the Law of Sines. You can also use it upside-down.

What good is it? The Law of Sines can be used to solve the following types of oblique triangles… AAS or ASA – Triangles with 2 known angles and 1 side SSA – Triangles with 2 known sides and 1 angle opposite one of the sides With these types of triangles, you will almost always have enough information/data to fill out one of the fractions.

Solving a triangle Solving a triangle means finding all the side and angle measures of the triangle. Once you know 2 angles, you can subtract from 180° to find the 3rd. To avoid rounding error, use given data instead of computed data whenever possible. We will use Law of Sines and Law of Cosines to fine missing pieces.

Example 1 – AAS A B C a b c 45 50 =30 85 I’m given both pieces for sinA/a and part of sinB/b, so we start there. Once I have 2 angles, I can find the missing angle by subtracting from 180°. C =180° – 45° – 50° = 85° Cross multiply and divide to get

Cross multiply and divide to get 45 50 =30 85 A B C a b c Cross multiply and divide to get Make sure students can get the right answer with their calculator. We’ll repeat the process to find side c. Remember to avoid rounded values when computing. We’re done when we know all 3 sides and all 3 angles.

Example 2 – ASA Let A = 35°, B = 10°, and c = 45 A B C a b c Since we can’t start one of the fractions, we’ll start by finding C. C = 180° – 35° – 10° = 135° 135 36.5 11.1 35 10 = 45 Since the angles were exact, this isn’t a rounded value. We use sinC/c as our starting fraction. Check for calculator ability Cross multiply and divide Using your calculator

Example 3 – SSA C Let A = 40°, b = 10, and a = 7 a b B A c We have enough information to use sinA/a as our starting fraction and can go to work on B a b 7 10 40 66.7 B A c We have to get rid of the sin function to find the answer for angle B. To do this, we apply the inverse operation. Check caculator We cross multiply and divide to get With a calculator

A B C a b c We need something to start the fraction sinC/c. Since it is critical that the angles total 180°, we will find C by subtracting from 180°. C = 180° – 40° – 66.7° = 73.3° 73.3 40 10 7 66.7 10.4 Now we use our starting fraction, sinA/a with sinC/c. Cross multiply and divide to finish. Check calculator

A Pain in the Angle Side Side Let’s consider the case where we have an angle, an adjacent side, and an opposite side. For example, I have angle A, side b, and side a. Sometimes a is too short to reach. You attempt to work this problem like example 3. Your calculator will give you an error message and catch the error. A b a Sometimes a is just right. It reaches with a right angle. You work this problem like example 3 and will never know there might have been a problem. A b a

Sometimes a is so long it only reaches one way Sometimes a is so long it only reaches one way. This problem also works just like example 3. You’ll never know this might have been difficult. A b a No triangle on this side Sometimes a is just the right length that it can form 2 different triangles. Following example 3 solves the outer triangle. You have to be on the look-out to catch the second triangle. a A b If you don’t get an error and a < b, then there will be either a right angle or 2 two triangles.

How do you know you have two triangles? SSA is given The angle given is acute The side opposite the angle is shorter than the other side given

Another Perspective on Starting the Second Triangle If the 2 sides are equal, the 2 angles are equal. Since supplementary angles total 180°, B’ = 180° – B. C’ Since A, b, and a are given, once you know B’ you can finish the triangle on the left. a a b B’ = 180 – B B B A c’

An Example of a Pain in the Angle Side Side Let A = 40°, b = 10, and a = 9. Angle side side C We will proceed like example 3. 40 b = 10 We have enough information for A and a start for B a = 9 A B c Cross multiply and divide to get

To get to angle B, you must unlock sin using the inverse. 40 b = 10 94.4 a = 9 45.6 A B c =14.0 Once you know 2 angles, find the third by subtracting from 180°. C = 180° – 40° – 45.6° = 94.4° Cross multiply and divide. Then use a calculator. We’re ready to look for side c. Remember to use the starter fraction (A)

Finding the Second Triangle Start by finding B’ = 180 – B Now solve this triangle. b = 10 a = 9 A B’ C’ b = 10 a = 9 40 134.4 c’ 40 B’ B = 45.6 A B’ = 180 – 45.6 = 134.4

Start by finding C’ = 180° – 40° – 134.4° C’ = 5.6 Finish by finding side c, using your starter fraction (A) 5.6 =1.4

A New Way to Find Area A B C a b c We all know that A = ½ bh. And a few slides back we found this. It looks like the base is c. If we drop the altitude in a different direction, we can add Making a little substitution gives me It’s always the 2 sides and the included angle.

An Area Example Find the area of a triangle with side a = 10, side b = 12, and angle C = 40. Choose the appropriate area formula. In this case we choose Area = ½ ab sinC. Fill in the blanks. Area = ½ (10)(12)sin40 Plug it into your calculator Area = 38.6 square units