1. LAW of SINES

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.
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
2. Two sides and an angle opposite one of them (SSA) C c a
3. Three sides (SSS) a c b c a B
4. Two sides and their included angle (SAS)
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
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.
4.15 ft
110
0.83 ft
Use the Law of Sines to find side b and c.
Example: Law of Sines - ASA

6. 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
21.26
48.74
48.23 in
C  180 – 110 – 48.74
= 21.26
Example: Single Solution Case - SSA

7. 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.
Example: No-Solution Case - SSA

8. Law of Sines: The Ambiguous Case
Given:
lengths of two sides and the angle opposite one of them (S-S-A)
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9. Possible Outcomes Case 1: If A is acute and a > b C b a
One SOLUTION B A c
Solve normally using the Law of Sines

10. Possible Outcomes Case 2: If A is acute and a < b
a. If a < b sinA a C b a
h = b sin A b B A h c A B c
NO SOLUTION

11. Possible Outcomes Case 2: If A is acute and a < b b. If a = b sinA
h = b sin A b h
= a A c B
1 SOLUTION

12. Possible Outcomes Case 2: If A is acute and a < b
h = b sin A
b. If a > b sinA C b a h a A B B c
2 SOLUTIONS

13. Possible Outcomes Case 3: If A is obtuse and a > b C a b A c B
ONE SOLUTION

14. Possible Outcomes Case 3: If A is obtuse and a ≤ b C a b A c B
NO SOLUTION

15. EXAMPLE 1 Given: ABC where a = 22 inches b = 12 inches a>b
mA = 42o
a>b
mA > mB
SINGLE–SOLUTION CASE
(acute)
Find m B, m C, and c.

16. sin A = sin B a b Sin B  0.36498 mB = 21.41o or 21o
Sine values of supplementary angles are equal.
The supplement of B is B2.  mB2=159o

17. mC = 180o – (42o + 21o) mC = 117o
sin A = sin C a c
c = inches
SINGLE–SOLUTION CASE

18. EXAMPLE 2 Given: ABC where c = 15 inches b = 25 inches c < b
mC = 85o
c < b
15 < 25 sin 85o
c ? b sin C
NO SOLUTION CASE
(acute)
Find m B, m C, and c.

19. sin A = sin B a b Sin B  1.66032 mB = ? Sin B > 1 NOT POSSIBLE !
Recall: – 1  sin   1
NO SOLUTION CASE
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20. EXAMPLE 3 Given: ABC where b = 15.2 inches a = 20 inches b < a
mB = 110o
b < a
NO SOLUTION CASE
(obtuse)
Find m B, m C, and c.

21. sin A = sin B a b Sin B  1.23644 mB = ? Sin B > 1 NOT POSSIBLE !
Recall: – 1  sin   1
NO SOLUTION CASE
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22. EXAMPLE 4 Given: ABC where a = 24 inches b = 36 inches a < b
mA = 25o
a < b
a ? b sin A
24 > 36 sin 25o
TWO – SOLUTION CASE
(acute)
Find m B, m C, and c.

23. sin A = sin B a b Sin B  0.63393 mB = 39.34o or 39o
The supplement of B is B2.  mB2 = 141o
mC1 = 180o – (25o + 39o) mC1 = 116o
mC2 = 180o – (25o+141o) mC2 = 14o

24. sin A = sin C a c1 c1 = 51.04 inches sin A = sin C a c2
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25. EXAMPLE 4 Final Answers: mB1 = 39o mC1 = 116o c1 = 51.04 in.
TWO – SOLUTION CASE
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26. Find m B, m C, and c, if they exist.
PRACTICE
Find m B, m C, and c, if they exist.
 1) a = 9.1, b = 12, mA = 35o
 2) a = 25, b = 46, mA = 37o
3) a = 15, b = 10, mA = 66o  
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27. Answers: 1)Case 1: mB=49o,mC=96o,c=15.78 Case 2:
2)No possible solution.
3)mB=38o,mC=76o,c=  
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28. Area of a Triangle The area A of a triangle is
where b is the base and h is an altitude drawn to that base

29. Area of SAS Triangles
If we know two sides a and b and the included angle C, then
The area A of a triangle equals one-half the product of two of its sides times the sine of their included angle.
Note: bsinC=h

30. Area of SAS Triangles Example.
Find the area A of the triangle for which a = 12, b = 15 and C = 52o