This document discusses the Law of Sines and Law of Cosines, which can be used to solve for missing sides and angles of oblique triangles (triangles without right angles). The Law of Sines relates the ratios of sides to opposite angles, while the Law of Cosines relates sides and angles. Several examples show how to apply these laws to find missing measurements in triangles given certain known values. The area of oblique triangles can also be found using these formulas.

1. 6.1 LAW OF SINES
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6.2 LAW OF COSINES

2. 2
Introduction

3. 3
Introduction
Oblique triangles—triangles that have no right angles.
Law of Sine
1. Two angles and any side (AAS or ASA)
2. Two sides and an angle opposite one of them
(SSA)
Law of Cosine
3. Three sides (SSS)
4. Two sides and their included angle (SAS)

4. 4
Introduction

5. 5
Introduction

6. 6
Example 1 – Given Two Angles and One Side—AAS
For the triangle in figure, C = 120, B = 29, andb = 28 feet.
Find the remaining angle and sides.
Solution:
The third angle of the triangle is
A = 180 – B – C
= 180 – 29 – 102
= 49. .

7. 7
Example 1 – Solution
The third angle of the triangle is
A = 180 – B – C
= 180 – 29 – 102
= 49.
By the Law of Sines, you have
.

8. 8
The Ambiguous Case (SSA)

9. 9
The Ambiguous Case (SSA)

10. 10
Example 3 – Single-Solution Case—SSA
For the triangle in Figure 6.4, a = 22 inches, b = 12 inches,
and A = 42. Find the remaining side and angles.
One solution: a  b
Figure 6.4

11. 11
Example 3 – Solution
By the Law of Sines, you have
Reciprocal form
Multiply each side by b.
Substitute for A, a, and b.
B is acute.

12. 12
Example 3 – Solution
Now, you can determine that
C  180 – 42 – 21.41
= 116.59.
Then, the remaining side is
cont’d

13. 13
Area of an Oblique Triangle

14. 14
Area of an Oblique Triangle
Area = (base)(height) = (c)(b sin A) = bc sin A.
A is acute. A is obtuse.

15. 15
Area of an Oblique Triangle

16. 16
Example 6 – 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 angle
C = 102, as shown in figure.
Then, the area of the triangle is
Area = ab sin C
= (90)(52)(sin 102)
 2289 square meters.

17. 17
Example 1 – Three Sides of a Triangle—SSS
Find the three angles of the triangle in Figure 6.11.
Solution:
It is a good idea first to find the angle opposite the longest
side—side b in this case. Using the alternative form of the
Law of Cosines, you find that
Figure 6.11

18. 18
Example 1 – Solution
Because cos B is negative, you know that B is an obtuse
angle given by B  116.80.
At this point, it is simpler to use the Law of Sines to
determine A.
cont’d

19. 19
Example 1 – Solution
You know that A must be acute because B is obtuse, and
a triangle can have, at most, one obtuse angle.
So, A  22.08 and
C  180 – 22.08 – 116.80
= 41.12.
cont’d

20. 20
Introduction
Do you see why it was wise to find the largest angle first in
Example 1? Knowing the cosine of an angle, you can
determine whether the angle is acute or obtuse. That is,
cos  > 0 for 0 <  < 90
cos  < 0 for 90 <  < 180.
So, in Example 1, once you found that angle B was obtuse,
you knew that angles A and C were both acute.
If the largest angle is acute, the remaining two angles are
acute also.
Acute
Obtuse

21. 21
Application

22. 22
Example 7 – An Application of the Law of Sines
The course for a boat race starts at point A in Figure 6.9
and proceeds in the direction S 52 W to point B, then in
the direction S 40 E to point C, and finally back to A. Point
C lies 8 kilometers directly south of point A. Approximate
the total distance of the race course.
Figure 6.9

23. 23
Example 7 – Solution
Because lines BD and AC are parallel, it follows that
BCA  CBD.
Consequently, triangle ABC has the measures shown in
Figure 6.10.
The measure of angle B is
180 – 52 – 40 = 88.
Using the Law of Sines,
Figure 6.10

24. 24
Example 7 – Solution
Because b = 8,
and
The total length of the course is approximately
Length  8 + 6.308 + 5.145
=19.453 kilometers.
cont’d