1. Law of Sines
Day 65

2. What You Should Learn
Use the Law of Sines to solve oblique triangles (AAS or ASA).
Use the Law of Sines to solve oblique triangles (SSA).
Find the areas of oblique triangles.
Use the Law of Sines to model and solve real-life problems.

3. Plan for the day When to use law of sines and law of cosines
Applying the law of sines
Law of sines - Ambiguous Case
Homework
6.1 page 416 #3, 5, all, 25 – 37 odd

4. Introduction
In this section, we will solve oblique triangles – triangles that have no right angles.
As standard notation, the angles of a triangle are labeled A, B, and C, and their opposite sides are labeled a, b, and c.
To solve an oblique triangle, we need to know the measure of at least one side and any two other measures of the triangle—either two sides, two angles, or one angle and one side.

5. Introduction This breaks down into the following four cases:
Two angles and any side (AAS or ASA)
Two sides and an angle opposite one of them (SSA)
Three sides (SSS)
Two sides and their included angle (SAS)
The first two cases can be solved using the Law of Sines, whereas the last two cases require the Law of Cosines.

6. Introduction
The Law of Sines can also be written in the reciprocal form:

7. Given Two Angles and One Side – AAS
For the triangle below C = 102, B = 29, and b = 28 feet. Find the remaining angle and sides.

8. Example AAS - Solution The third angle of the triangle is
A = 180 – B – C
= 180 – 29 – 102
= 49.
By the Law of Sines, you have .

9. Example AAS – Solution
cont’d
Using b = 28 produces
and

10. Law of Sines For non right triangles Law of sines Try this: A B C c a

11. Let’s look at this: Example 1
Given a triangle, demonstrate using the Law of Sines that it is a valid triangle (numbers are rounded to the nearest tenth so they may be up to a tenth off):
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
Is it valid??

12. And this: Example 2
Given a triangle, demonstrate using the Law of Sines that it is a valid triangle (numbers are rounded to the nearest tenth so they may be up to a tenth off):
a = 5 A = 40o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
Is it valid??
Why does this work?

13. Looking at these two examples
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
In both cases a, b and A are the same (two sides and an angle) but they produced two different triangles
Why??

14. Here is what happened a = 5 A = 40o b = 7 B = c = C =
What is sin 64.1?
What is sin 115.9?
What is the relationship between these two angles?
Remember the sine of an angle in the first quadrant (acute: 0o – 90o) and second quadrant, (obtuse: 90o – 180o)are the same!

15. The Ambiguous Case (SSA)

16. The Ambiguous Case (SSA)
In our first example we saw that two angles and one side determine a unique triangle.
However, if two sides and one opposite angle are given, three possible situations can occur:
(1) no such triangle exists,
(2) one such triangle exists, or
(3) two distinct triangles may satisfy the conditions.

17. Back to these examples Given two sides and an angle across
a = 5 A = 40o
b = 7 B = 64.1o
c = 7.5 C = 75.9o
b = 7 B = 115.9o
c = 3.2 C = 24.1o
The Ambiguous Case

18. The Ambiguous Case (SSA)

19. Ambiguous Case Is it Law or Sines or Law of Cosines
Law of Cosines – solve based upon one solution
Law of Sines – go to #2
Law of Sines - Is it the SSA case? (Two sides and angle opposite)
No – not ambiguous, solve based upon one solution
Yes – go to #3.
Is the side opposite the angle the shortest side?
Yes – go to #4
Is the angle obtuse?
No – go to #5
Yes – no solution
Calculate the height of the triangle height = the side not opposite the angle x the sine of the angle
If the side opposite the angle is shorter than the height – no solution
If the side opposite the angle is equal to the height – one solution
If the side opposite the angle is longer than the height – two solutions

20. How many solutions are there?
A = 30o a = 5 b = 3
B = 50o a = 6 b = 5
C = 80o b = 5 c = 6
A = 40o a = 4 b = 8
a = 5 b = 4 c = 6
B = 20o a = 10 c = 15
A = 25o a = 3 b = 6
C = 75o a = 5 b = 3 1 2 1 1 1 2 1

21. Example – Single-Solution Case—SSA
For the triangle below, a = 22 inches, b = 12 inches, and A = 42. Find the remaining side and angles.
One solution: a  b

22. Example – Solution SSA By the Law of Sines, you have Reciprocal form
Multiply each side by b.
Substitute for A, a, and b.
B is acute.

23. Example – Solution SSA Now, you can determine that
cont’d
Now, you can determine that
C  180 – 42 – 21.41
= .
Then, the remaining side is

24. What about more than one solution?

25. How many solutions are there?
A = 30o a = 5 b = 3
B = 50o a = 6 b = 5
C = 80o b = 5 c = 6
A = 40o a = 4 b = 8
a = 5 b = 4 c = 6
B = 20o a = 10 c = 15
A = 25o a = 3 b = 6
C = 75o a = 5 b = 3 1 2
Solve #2, 7

26. Area of an Oblique Triangle

27. Area of an Oblique Triangle
The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle.
Referring to the triangles below, that each triangle has a height of h = b sin A.
A is acute.
A is obtuse.

28. Area of a Triangle - SAS
SAS – you know two sides: b, c and the angle between: A
Remember area of a triangle is ½ base ● height
Base = b
Height = c ● sin A
 Area = ½ bc(sinA) A B C c a b h
Looking at this from all three sides:
Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

29. Area of an Oblique Triangle

30. Example – Finding the Area of a Triangular Lot
Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102.
Solution:
Consider a = 90 meters, b = 52 meters, and the included angle C = 102
Then, the area of the triangle is
Area = ½ ab sin C
= ½ (90)(52)(sin102)
 2289 square meters.

31. Homework 35
Section 6.1 p. 416:
3, 5, all, odd