1. Law of Sines

2. Objectives
Use the Law of Sines to solve oblique triangles (AAS or ASA). Use the Law of Sines to solve oblique triangles (SSA). Find the areas of oblique triangles. Use the Law of Sines to model and solve real-life problems.

3. Plan for the day
When to use law of sines
Applying the law of sines

4. Introduction
In this section, we will solve oblique triangles – triangles that have no right angles. As standard notation, the angles of a triangle are labeled A, B, and C, and their opposite sides are labeled a, b, and c. To solve an oblique triangle, we need to know the measure of at least one side and any two other measures of the triangle—either two sides, two angles, or one angle and one side.

5. Introduction This breaks down into the following four cases:
Two angles and any side (AAS or ASA)
Two sides and an angle opposite one of them (SSA)
Three sides (SSS)
Two sides and their included angle (SAS)
The first two cases can be solved using the Law of Sines, whereas the last two cases require the Law of Cosines.

6. Introduction
The Law of Sines can also be written in the reciprocal form:

7. Given Two Angles and One Side – AAS
For the triangle below m∠C = 102, m∠B = 29, and b = 28 feet. Find the remaining angle and sides.
49°
The third angle of the triangle is:
m∠A = 180 – m∠B – m∠C
m∠A = 180 – 29 – 102
m∠A = 49

8. 28 sin 29° = 𝑎 sin 49° 28 sin 29° = 𝑐 sin 102°
28 * sin 49° = a * sin 29°
sin 29°
sin 29°
28 * sin 102° = c * sin 29°
sin 29°
sin 29°
a ≈ ft.
c ≈ ft.

9. Try this: m∠A = 43°, m∠B = 67°, c = 45 mm
Law of Sines A B C c a b
For non-right triangles
Law of sines
Try this: m∠A = 43°, m∠B = 67°, c = 45 mm 45
67°
43°
70°
The third angle of the triangle is:
m∠C = 180 – m∠A – m∠B
m∠C = 180 – 43 – 67
m∠C = 70

10. 45 67° 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶 43° 70° 𝑏 sin 67° = 45 sin 70°A B C c a b 45
67°
𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶
43°
70°
𝑏 sin 67° = 45 sin 70°
𝑎 sin 43° = 45 sin 70°
b * sin 70° = 45 * sin 67°
sin 70°
sin 70°
a * sin 70° = 45 * sin 43°
sin 70°
sin 70°
b ≈ mm
a ≈ mm

11. Example – Single-Solution Case—SSA
For the triangle below, a = 22 inches, b = 12 inches, and m∠A = 42. Find the remaining side and angles.
𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶
22 sin 42° = 12 sin 𝐵
sin B = some number
22 * sin B = sin (42) * 12
Push 2nd SIN 22 22
Push 2nd (–) for ANS
m∠B ≈ 21.40°
𝑠𝑖𝑛 −1 (ANS) ENTER

12. Law of Sines The third angle of the triangle is:
116.6°
21.4°
The third angle of the triangle is:
m∠C = 180 – m∠A – m∠B
m∠C = 180 – 42 – 21.4
m∠C = 116.6
22 * sin = c * sin (42)
22 sin 42° = 𝑐 sin 116.6
sin 42
sin 42
c ≈ 29.40

13. Law of Sines Worksheet #1-17 odd

14. Law of Cosines

15. Objectives:
Use the Law of Cosines to solve oblique triangles (SSS or SAS).
Use the Law of Cosines to model and solve real-life problems.

16. Warm-up:
Solve for side a in ∆ABC when m∠A = 29°, m∠B = 62° and c = 11.5 A B C c a b
The third angle of the triangle is:
m∠C = 180 – m∠A – m∠B
m∠C = 180 – 29 – 62
11.5
62°
m∠C = 89
29°
89°
11.5 sin 89 = 𝑎 sin 29
11.5 * sin 29 = a * sin 89
sin 89
sin 89
11.5 sin 89 = 𝑏 sin 62
b ≈ 10.16
a ≈ 5.58

17. Introduction Four cases. Two angles and any side (AAS or ASA)
Two sides and an angle opposite one of them (SSA)
Three sides (SSS)
Two sides and their included angle (SAS)
The first two cases can be solved using the Law of Sines, whereas the last two cases require the Law of Cosines.

18. Law of Sines 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶 For non right trianglesB C c a b
𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶

19. Law of Cosines: Introduction
Two cases remain in the list of conditions needed to solve an oblique triangle – SSS and SAS.
If you are given three sides (SSS), or two sides and their included angle (SAS), none of the ratios in the Law of Sines would be complete.
In such cases, you can use the Law of Cosines.

20. Law of Cosines A B C c a b
Side, Angle, Side

21. 1. m∠A = 52o b = 6 c = 8 a2 = b2 + c2 – 2bc cos A
Try these: Solve ∆ ABC. Round angle measures to the nearest degree and side measures to the nearest tenth.
1. m∠A = 52o b = 6 c = 8 A B C c a b
a2 = b2 + c2 – 2bc cos A 8
a2 = – 2(6)(8) cos 52
52°
a2 = – 96 cos 52
a2 = 100 – 96 cos 52 6
a ≈ 6.4

22. 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶 6.4 sin 52 = 6 sin 𝐵 m∠C ≈ 80 8 6.4 48°A B C c a b 8
6.4
48°
6.4 sin 52 = 6 sin 𝐵
52°
6.4 * sin B = 6 * sin 52 6
6.4
6.4
m∠B ≈ 48°
sin B = some number
Push 2nd SIN
The third angle of the triangle is:
m∠C = 180 – m∠A – m∠B
m∠C = 180 – 52 – 48
Push 2nd (–) for ANS
𝑠𝑖𝑛 −1 (ANS) ENTER
m∠C ≈ 80

23. 2. m∠B = 58o a = 9 c = 14 b2 = a2 + c2 – 2ac cos B
58° 9
b2 = – 2(9)(14) cos 58
12.0
b2 = – 252 cos 58
b2 = 277 – 252 cos 58
b ≈ 12.0

24. 2. m∠B = 58° a = 9 c = 14 A B C c a b
12 sin 58 = 9 sin 𝐴 14
58° 9
12 * sin A = 9 * sin 58 12 12
12.0
m∠A ≈ 39°
sin C = some number
Push 2nd SIN
The third angle of the triangle is:
m∠C = 180 – m∠A – m∠B
m∠C = 180 – 39 – 58
m∠C ≈ 83
Push 2nd (–) for ANS
𝑠𝑖𝑛 −1 (ANS) ENTER

25. Law of Cosines
Side, Side, Side A B C c a b

26. Law of Cosines cos A = 𝑏 2 + 𝑐 2 − 𝑎 2 2𝑏𝑐 cos B = 𝑎 2 + 𝑐 2 − 𝑏 2 2𝑎𝑐
SSS
cos B = 𝑎 2 + 𝑐 2 − 𝑏 2 2𝑎𝑐
cos C = 𝑎 2 + 𝑏 2 − 𝑐 2 2𝑎𝑏

27. Try these: Solve ∆ ABC. Round angle measures to the nearest degree.
10.3 21
cos A = 𝑏 2 + 𝑐 2 − 𝑎 2 2𝑏𝑐
cos A = − (16.7)(10.3)
16.7
cos A = −

28. A B C c a b
cos A = −
10.3 21
cos A = −
99°
m∠A ≈ 99°
16.7
The third angle of the triangle is:
m∠C = 180 – m∠A – m∠B
m∠C = 180 – 99 – 52
21 sin 99 = sin 𝐵
21 * sin B = * sin 99
m∠C ≈ 29° 21 21
m∠B ≈ 52°

29. 4. a = b = c = 21.8 A B C c a b
21.8 19
cos A = 𝑏 2 + 𝑐 2 − 𝑎 2 2𝑏𝑐
cos A = − (24.3)(21.8)
16.7
cos A = −

30. A B C c a b
cos A = −
21.8 19
cos A =
48°
m∠A ≈ 48°
24.3
The third angle of the triangle is:
m∠C = 180 – m∠A – m∠B
m∠C = 180 – 48 – 59
19 sin 48 = sin 𝐵
19 * sin B = * sin 48
m∠C ≈ 75° 19 21
m∠B ≈ 59°

31. Applications

32. Applications 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?

33. Solution
In triangle HPF, m∠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 h2 = – 2(43)(60) cos 45. h2  So, the approximate distance from the pitcher’s mound to first base is h  feet.

34. Applications of the Law of Cosines
The leading edge of each wing of the B-2 Stealth Bomber measures feet in length.
The angle between the wing's leading edges is °.
What is the wing span (the distance from A to C)? B
105.6 ft

35. C A
105.6
109.05° B a b c
b2 = a2 + c2 – 2ac cos B
b2 = – 2(105.6)(105.6) cos
b = ft

36. Area of an Oblique Triangle

37. 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.

38. Area of a Triangle - SAS
SAS – you know two sides: b, c and the angle between: A
Remember area of a triangle is base ● height
Base = b
Height = c ● sin A
 Area = bc(sinA) A B C c a b h
Looking at this from all three sides:
Area = ab(sin C) = ac(sin B) = bc (sin A)

39. Area of an Oblique Triangle

40. 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 = ab sin C
= (90)(52)(sin102)
 2289 square meters.

41. Heron’s Formula

42. 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.).

43. Area of a Triangle Law of Cosines Case - SSSB C c a b h
SSS – Given all three sides
Heron’s formula:
where s = 𝑎 + 𝑏 +𝑐 2

44. 1. Given the triangle with three sides of 6, 8, 10, find the area.
A = 24 units2

45. 2. Given the triangle with three sides of 12, 15, 21, find the area
A = 7776 units2