Math (Grade 9) Unit IV - Law of Sines and Law of Cosines of a Triangle

1. Oblique Triangle
An Oblique Triangle is a non-right
triangle.

2. Oblique Triangle
• There are several laws that can be use to solve
oblique triangle. These are the law of sines,
law of cosines and law of tangents.
• As in solving right triangles, you should know
three parts of an oblique triangle to find the
other three missing parts.

3. Oblique Triangle
Four Cases
1. ASA or SAA – Law of Sines
2. SSA – law of Sines ( ambiguous case)
3. SAS – Law of cosines
4. SSS- Law of Cosines

4. Oblique Triangle
1. Given: A, b, C Law of Sin
2. c, B, C Law of Sin
3. c, a, C Law of Sin
4. b,A,c Law of Cos
5. c, B, a Law of cos
6. a, b, c Law of cos

5. Oblique Triangle
Law of cos
Law of sin
Law of cos

6. Law of Sines
• If A, B, and C, are the angles of any triangle,
and a,b, and c, are respectively, the measures
of the sides opposite these angles, then
𝑎
sin 𝐴
=
𝑏
sin 𝐵
=
𝑐
sin 𝐶

7. Law of Sines
𝑎
sin 𝐴
=
𝑏
sin 𝐵
𝑏
sin 𝐵
=
𝑐
sin 𝐶
𝑎
sin 𝐴
=
𝑐
sin 𝐶

8. Law of Sines
1. Solve the triangle given.
Solution: A + B + C = 180
B = 180 – 51.2 – 48.6
B = 80.2o
From the law of sine
23.5
sin 51.2
=
𝑏
sin 80.2
b=
23.5(𝑠𝑖𝑛80.2)
𝑠𝑖𝑛51.2
b= 29.7
80.2o
29.7

9. Law of Sines
1. Solve the triangle given.
B= 80.2
b = 29.7
From the law of sine
23.5
sin 51.2
=
𝑐
sin 48.6
c=
23.5(𝑠𝑖𝑛48.6)
𝑠𝑖𝑛51.2
c= 22.6
22.6
80.2o
29.7

10. Solve the triangle ABC, given a=62.5, A=112o,
and C=42
B=180-112-42
B=26
62.5
sin 112
=
𝑐
sin 42
c=
62.5(𝑠𝑖𝑛42)
𝑠𝑖𝑛112
c= 45.1

11. Solve the triangle ABC, given a=62.5, A=112o, and C=42
B=180-112-42
B=26
62.5
sin 112
=
𝑏
sin 26
b=
62.5(𝑠𝑖𝑛26)
𝑠𝑖𝑛112
b= 29.5
Answers B=26o b= 45.1 c=29.5

12. If
𝑠𝑖𝑛𝐵
𝑏
=
sin 𝐶
𝑐
, then B =____
a) B = 𝑠𝑖𝑛−1 𝑏𝑠𝑖𝑛𝐶
𝑐
b) B = 𝑠𝑖𝑛−1 𝑐𝑠𝑖𝑛𝐶
𝑏
c)B = 𝑠𝑖𝑛−1 𝑏
𝑐𝑠𝑖𝑛 𝐶
Ans. a

13. If
𝑎
sin 𝐴
=
𝑏
sin 𝐵
, then a =____
a) a =
𝑏𝑠𝑖𝑛𝐵
sin 𝐴
b) a =
𝑠𝑖𝑛𝐵
bsin 𝐴
c) a =
𝑏𝑠𝑖𝑛𝐴
sin 𝐵
Ans. c
If
𝑎
sin 𝐴
=
𝑐
sin 𝐶
, then c =____
a)
𝑎𝑠𝑖𝑛𝐴
sin 𝐶
= 𝑐 b)
𝑎𝑠𝑖𝑛𝐶
sin 𝐴
= 𝑐 c)
𝑠𝑖𝑛𝐶
asin 𝐵
= 𝑐
Ans. B

14. Oblique Triangle
Give the appropriate law.
1. SAS –
Law of Cos
2. SSS –
Law of cos
3. SSA –
law of Sines ( ambiguous case)
4. ASA -
Law of sin
5. SAA –
law of sin

15. Law of Cosines
• For any Triangle ABC with sides a,b, and c,
Use to solve the missing sides
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C

16. Law of Cosines
• For any Triangle ABC with sides a,b, and c,
Use to solve the missing sides
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C

17. Law of Cosines
• For any Triangle ABC with sides a,b, and c,
Use to solve the missing angles
cos A=
b2 + c2 − a2
2𝑏𝑐
cos B=
a2 + c2 − b2
2𝑎𝑐
cos C=
a2 + b2 − c2
2𝑎𝑏

18. Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
a2 = b2 + c2 – 2bc cos A
a2 = 12 + 32 – 2(1)(3) cos 80
a2 = 1 + 9 – 6cos 80
a2 = 10 – 1.04
a2 = 8.96
a= 8.96
a =2.99
a=2.99

19. Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
sin 𝐴
𝑎
=
sin 𝐵
𝑏
sin 80
2.99
=
sin 𝐵
1
(1 )sin 80
2.99
= sin 𝐵
0.3294 = sin B
𝑠𝑖𝑛−1
0.3294 = 𝐵
19.2o = B
a=2.99
19.2o

20. Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
Since A+B+C = 180o
Then 80o +19.2o +C=180o
99.2o +C =180o
C =180o -99.2o
C = 80.8o
a=2.99
19.2o
80.8o

21. Law of
• Solve the triangle with a = 5, b = 8, and c=9
a2 = b2 + c2 – 2bc cos A
52 = 82 + 92 – 2(8)(9) cos A
25=64+81-144cosA
25=145-144cosA
144cosA=145-25
cosA=120/144
cosA = 0.8333
A= 𝑐𝑜𝑠−1
0.8333
A = 33.6o
33.6o

22. Law of Cos
• Solve the triangle with a = 5, b = 8, and c=9
sin 𝐴
𝑎
=
sin 𝐵
𝑏
sin 33.6
5
=
sin 𝐵
8
(8 )sin 33.6
5
= sin 𝐵
0.8854 = sin B
𝑠𝑖𝑛−1
0.8854 = 𝐵
62.3o = B
33.6o62.3o

23. Law of Cos
• Solve the triangle with a = 5, b = 8, and c=9
Since A+B+C = 180o
33.6o +62.3o +C=180o
95.9o +C =180o
C =180o -95.9o
C = 84.1o
33.6o62.3o
84.1o

24. • A Triangular lot has dimensions 20.6m, 31.4m,
and 38.3m. Find the angles at the corners of the
property.
20.62=31.42+38.32 -2(31.4)(38.3)cosA
424.36=2452.85-2405.24cosA
cosA=2028.49/2405.24
cosA=0.8434
A= 32.5o

25. • A Triangular lot has dimensions 20.6m, 31.4m,
and 38.3m. Find the angles at the corners of
the property.
sin 32.5/20.6 = sin B/31.4
0.8190 =sin B
B = 54. 98o or 55
C =180 – 32.5 – 54.98 = 92.52o

26. What is the length of side b?
b=3.08 C=79

27. • What is the size of Angle C?
C=40.51

28. • What is the size of Angle P?

29. The diagram shows part of a logo design.
There is one known angle of 142°.
Calculate the sizes of the other two angles.

30. • Mrs Jones goes on a round trip from Town A
to Town B to Town C and back to Town A, as
shown in the following diagram. All roads are
straight. To the nearest mile. How long is the
round trip?

31. • What is the length of side c?
c2=5.32+3.62-2(5.3)(3.6)cos59
c=4.63

32. • Find angle A
82=52+92-2(5)(9)cosA
A=62.18

33. • Find angle B

34. • Ayton is 25 miles due north of Beeton. Ceeton lies to the east
side the road joining Ayton to Beeton, and is 47 miles from
Ayton and 63 miles from Beeton. (All roads are straight.)
Calculate the three-figure bearing of Ceeton from Ayton.
Note A three-figure bearing is always measured in a clockwise
sense from the direction North.