This document discusses oblique triangles and the sine law and cosine law for solving triangles. It provides examples of applying these laws to find missing angles and sides of triangles given various parts of triangles, such as two sides and the included angle, two angles and the included side, or three sides. The seven examples show step-by-step workings of applying the sine and cosine laws to find the missing parts in different triangle scenarios.

2. •Oblique Triangles
•Sine Law
•Cosine Law
•Application of the Sine
and the Cosine Law

3. •Oblique Triangles
a. The Sum of the three angles is equal to 1800 .A
+ B + C = 1800;
b. The greater angles subtends the longer side, if
A> B > C, then a> b> c;
c. A triangle is determined or fixed in form and
in size if the following parts are given:
a. Two angles and the side opposite one of
them;
b. Two sides and the angle opposite one of
them;
c. Two sides and the included angle;
d. Two angles and the included side;
e. Three sides
A plane triangle, which does
not contain a right angle
Figure 1
A
c
a
b
C
B
A
b
ac
C
B

4. SINE LAW
Each side of a triangle is directly
proportional to the sine of
the opposite angle.
A b
ac
C
B
A
c
a
b
C
B

5. In its common form, the Sine
Law is written as:
C
c
B
b
A
a
sinsinsin
c
C
b
B
a
A sinsinsin
The above equation can also be written in
the following form:

6. Use the Sine Law when there is a
“pair” in the given data.
“Pair” means a side and its opposite
angle. For example, angle A and
side a constitutes a pair.
TIP

7. COSINE LAW
The square of any side of a triangle is equal to the
sum of the squares of the other two sides minus twice
the product of these two sides times the cosine of the
angle between these two sides.
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C

8. Taking the inverse cosine function
of both sides, gives:
bc
acb
A
2
cos
222
1
ac
bca
B
2
cos
222
1
ab
cba
C
2
cos
222
1
Similarly,

9. TIP
Use the COSINE
Law when there is
“no pair” in the
given data.

10. Find: C, b, and c
Example 1. Solve the triangle ABC,
given the following parts: a = 450,
A = 11020’, B = 43034’
Applications of the Sine Law and the Cosine Law:
A b
B
a=450
C
c 43034’
11020’

11. A + B + C = 1800
C = 1800 - A - B
= 1800 11020’ - 43034’
C = 125006’
Solution:
A
a
B
b
sinsin
'0
'0
2011sin
)3443(sin450
A
Ca
c
sin
sin
'0
'0
2011sin
)06125(sin450
A
a
C
c
sinsin
19652.0
)81815.0(450
44.873,1c
A
Ba
b
sin
sin
19652.0
)68920.0(450
16.578,1b

12. Find: C, b, and B
Example 2. Solve triangle
ABC, given the following
parts: a = 72.53, A =800
Applications of the Sine Law and the Cosine Law:

13. Solution:
a
A
c
C sinsin
a
Ac
C
sin
sin
53.72
80sin38.117 0
59378.1sinC ,since sinC > 1
the triangle is impossible.

14. Find: B, c, and C
Example 3. Solve the triangle
ABC, given the following parts:
a = 456, b = 321, A =570
Applications of the Sine Law and the Cosine Law:

15. Solution:
a
A
b
B sinsin
a
Ab
B
sin
sin
456
57sin321 0
59038.0sin B
59038.0sin 1
B
1840.36B

16. A
b = 321
B
a=456
C
c
570
A
a
c
c
sinsin
A
Ca
c
sin
sin
0
'0
57sin
4986sin456
88.542c
Sin B = 0.59038
B = sin-1 0.59038
B = 36011’
C = 1800 – A – B
C = 1800 – 570 – 36011’
C = 86049’

17. Find: C, a, and A
Example 4. Solve triangle
ABC, given the following parts:
b = 176, c = 215, B =360
Applications of the Sine Law and the Cosine Law:

18. Solution:
b
B
c
c sin
sin
b
Bc
C
sin
sin
176
36sin215 0
71803.0sinC
71803.0sin 1
C
8921.45C

19. A
b = 176
B
a
C
c =215
360
B
b
A
a
sinsin
B
Ab
a
sin
sin
0
'0
36sin
0698sin176
a
44.296a
If C is acute:
sin C = 0.71803
C = sin-1 0.71803
C = 45054’
A = 1800 – B – C
= 1800 – 360 – 45054’
A = 98006’

20. A
b = 176
B = 360
a
C
c =215
B
b
A
a
sinsin
B
Ab
a
sin
sin
0
'0
36sin
5490sin176
a
48.51a
C = 1800 - 45054’
C = 134006’
A = 1800 – B – C
A = 1800 – 360 - 134006’
A = 90054’

21. Find: b, A, and C
Given two sides and the included angle
Example 5. Solve triangle ABC,
given the following parts: a = 25,
c = 30, B =50005’
Applications of the Sine Law and the Cosine Law:

22. Solution:
b
B =50005’
a =25
C
c =30
A
b
B
a
A sinsin
b
Ba
A
sin
sin
72.23
0550sin25
sin
'0
A
80837.0sin A
'0
5653A
BAC 0
180
'0'00
05505653180C
'0
5975C
b2 = a2 + c2 – 2ac cos B
= 252 + 302 – 2(25)(30) cos 50005’
b2 = 562.49
49.562b
b = 23.72

23. Find: C, a, and B
Given two angles and the included aside
Example 6. Solve triangle ABC,
given the following parts: A = 370, B =
150, c =17
Applications of the Sine Law and the Cosine Law:

24. Solution:
b
B =150
a
C
c =17
C = 1800 – A – B
= 1800 – 370 - 150
C = 1280

25. C
c
A
a
sinsin
C
Ac
a
sin
sin
0
0
128sin
37sin17
a
98.12a
C
c
B
b
sinsin
C
Bc
b
sin
sin
0
0
128sin
15sin17
b
58.5b

26. Find: A, B and C
Given three sides
Example 7. Solve triangle ABC, given the
following parts: a = 4, b = 5, c =6
Applications of the Sine Law and the Cosine Law:
b = 5
B
a = 4
C
c =6
A

27. Solution: There is no “pair” in the given data,
so use the COSINE LAW
bc
acb
A
2
cos
222
1
)6)(5(2
465
cos
222
1
75000.0cos 1
'0
2541A

28. ac
bca
B
2
cos
222
1
)6)(4(2
564
cos
222
1
56250.0cos 1
'0
4655B
ab
cba
C
2
cos
222
1
)5)(4(2
654
cos
222
1
12500.0cos 1
'0
4982C

29. 1800 = A + B + C
1800 = 41025’ + 55046’ + 82049’
1800 = 180

30. End of Slides
Submitted
to:ALICIA F. ALINDOTE
Instructor