1. Law of Sines and Law of Cosines
Digital Lesson
Law of Sines and Law of Cosines

2. Definition: Oblique Triangles
An oblique triangle is a triangle that has no right angles. C B A a b c
To solve an oblique triangle, you need to know the measure of at least one side and the measures of any other two parts of the triangle – two sides, two angles, or one angle and one side.
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Definition: Oblique Triangles

3. Solving Oblique Triangles
The following cases are considered when solving oblique triangles.
Two angles and any side (AAS or ASA) A C c A B c
Law of sines
2. Two sides and an angle opposite one of them (SSA) C c a
Law of sines
3. Three sides (SSS) a c b
Law of cosines c a B
Law of
cosines
4. Two sides and their included angle (SAS)
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Solving Oblique Triangles

4. Definition: Law of Sines
The first two cases can be solved using the Law of Sines. (The last two cases can be solved using the Law of Cosines.)
Law of Sines
If ABC is an oblique triangle with sides a, b, and c, then C B A b h c a C B A b h c a
Acute Triangle
Obtuse Triangle
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Definition: Law of Sines

5. Example: Law of Sines - ASA
Example (ASA):
Find the remaining angle and sides of the triangle. C B A b c
60
10
a = 4.5 ft
The third angle in the triangle is A = 180 – C – B
= 180 – 10 – 60
= 110.
Use the Law of Sines to find side b and c.
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Example: Law of Sines - ASA

6. The Ambiguous Case (SSA) Law of Sines
In earlier examples, you saw that two angles and one side determine a unique triangle.
However, if two sides and one opposite angle are given, then three possible situations can occur: (1) no such triangle exists,
(2) one such triangle exists, or
(3) two distinct triangles satisfy the
conditions.

7. The Ambiguous Case (SSA)

8. Example 3 – Single-Solution Case—SSA
For the triangle in Figure 6.5, a = 22 inches, b = 12 inches, and A = 42 . Find the remaining side and angles.
One solution: a  b
Figure 6.5

9. Now you can determine that
Example 3 – Solution
cont’d
Check for the other “potential angle”
C  180  –  =
( = which is more than 180  so there
is only one triangle.)
Now you can determine that
C  180  – 42  –  = 
Multiply each side by b
Substitute for A, a, and b.
B is acute.

10. Multiply each side by sin C.
Example 3 – Solution
cont’d
Then the remaining side is given by
Law of Sines
Multiply each side by sin C.
Substitute for a, A, and C.
Simplify.

11. Example: Single Solution Case - SSA
Example (SSA):
Use the Law of Sines to solve the triangle.
A = 110, a = 125 inches, b = 100 inches C B A
b = 100 in c
a = 125 in
110
C  180 – 110 – 48.74
= 21.26
Since the “given angle is already obtuse, there will only be one triangle.
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Example: Single Solution Case - SSA

12. Example: No-Solution Case - SSA
Example (SSA):
Use the Law of Sines to solve the triangle.
A = 76, a = 18 inches, b = 20 inches C A B
b = 20 in
a = 18 in
76
There is no angle whose sine is
There is no triangle satisfying the given conditions.
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Example: No-Solution Case - SSA

13. Example: Two-Solution Case - SSA
Example (SSA):
a = 11.4 cm C A B1
b = 12.8 cm c
58
Use the Law of Sines to solve the triangle.
A = 58, a = 11.4 cm, b = 12.8 cm
C  180 – 58 – 72.2 = 49.8
Check for the other “potential angle”
C  180  – 72.2  = 
(107.8  + 58  =  which is less than 180  so there are two triangles.)
Example continues.
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Example: Two-Solution Case - SSA

14. Example: Two-Solution Case – SSA continued
Example (SSA) continued:
72.2
10.3 cm
49.8
a = 11.4 cm C A B1
b = 12.8 cm c
58
Use the Law of Sines to solve the second triangle.
A = 58, a = 11.4 cm, b = 12.8 cm
B2  180 – 72.2 = 
C  180 – 58 – 107.8 = 14.2 C A B2
b = 12.8 cm c
a = 11.4 cm
58
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Example: Two-Solution Case – SSA continued

15. Area of an Oblique Triangle

16. 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.

17. 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)

18. Area of an Oblique Triangle

19. 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.

20. Law of Cosines (Solving for a side)
(SSS and SAS) can be solved using the Law of Cosines.
(Solving for a side)
Always solve for the angle across from the longest side first!

21. Law of Cosines ( solving for an angle)
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22. Example: Law of Cosines - SSS
Find the three angles of the triangle. C B A 8 6 12
Find the angle opposite the longest side first.
Law of Sines:
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Example: Law of Cosines - SSS

23. Example: Law of Cosines - SASB A
6.2
75
9.5
Solve the triangle.
Law of Cosines:
Law of Sines:
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Example: Law of Cosines - SAS

24. Applications

25. An Application of the Law of Cosines
The pitcher’s mound on a women’s softball field is 43 feet from home plate and the distance between the bases is 60 feet (The pitcher’s mound is not halfway between home plate and second base.) How far is the pitcher’s mound from first base?

26. Solution
In triangle HPF, H = 45 (line HP bisects the right angle at H), f = 43, and p = 60.
Using the Law of Cosines for this SAS case, you have
h2 = f 2 + p2 – 2fp cos H
= – 2(43)(60) cos 45 
So, the approximate distance from the pitcher’s mound to first base is  feet.

27. Heron’s Formula

28. Heron’s Area Formula
The Law of Cosines can be used to establish the following formula for the area of a triangle. This formula is called Heron’s Area Formula after the Greek mathematician Heron (c. 100 B.C.).

29. Area of a Triangle Law of Cosines Case - SSSB C c a b h
SSS – Given all three sides
Heron’s formula:

30. Try these
Given the triangle with three sides of 6, 8, 10 find the area
Given the triangle with three sides of 12, 15, 21 find the area