1. Section 6.1 Law of Sines

2. 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

3. What we already know A B C a b c

4. 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…

5. 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.

6. 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.

7. 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.

8. 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

9. Cross multiply and divide to get45 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.

10. 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

11. 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

12. 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

13. 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

14. 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.

15. 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

16. 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’

17. 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

18. 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)

19. 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

20. 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

21. 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.

22. 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