1. The Law
of SINES

2. Working with Non-right Triangles
We wish to solve triangles which are not right triangles B A C a c b h
Drawing a different altitude will allow to prove similarly these both =

3. The Law of SINES
For any triangle (right, acute or obtuse), you may use the following formula to solve for missing sides or angles:

4. AAS - 2 angles and 1 adjacent side
Use Law of SINES when ...
you have 3 dimensions of a triangle and you need to find the other 3 dimensions - they cannot be just ANY 3 dimensions though, or you won’t have enough info to solve the Law of Sines equation. Use the Law of Sines if you are given:
AAS - 2 angles and 1 adjacent side
ASA - 2 angles and their included side
SSA (this is an ambiguous case)

5. Example 1
You are given a triangle, ABC, with angle A = 70°, angle B = 80° and side a = 12 cm. Find the measures of angle C and sides b and c.
* In this section, angles are named with capital letters and the side opposite an angle is named with the same lower case letter .*

6. Example 1 (con’t) A C B 70° 80° a = 12 c b
The angles in a ∆ total 180°, so angle C = 30°.
Set up the Law of Sines to find side b:

7. Example 1 (con’t) A C B 70° 80° a = 12 c b = 12.6 30°
Set up the Law of Sines to find side c:

8. Example 1 (solution) A C B 70° 80° a = 12 c = 6.4 b = 12.6 30°
Angle C = 30°
Side b = 12.6 cm
Side c = 6.4 cm
Note:
We used the given values of A and a in both calculations. Your answer is more accurate if you do not used rounded values in calculations.

9. Example 2
You are given a triangle, ABC, with angle C = 115°, angle B = 30° and side a = 30 cm. Find the measures of angle A and sides b and c.

10. Example 2 (con’t)
To solve for the missing sides or angles, we must have an angle and opposite side to set up the first equation.
We MUST find angle A first because the only side given is side a.
The angles in a ∆ total 180°, so angle A = 35°. A C B
115°
30°
a = 30 c b

11. Example 2 (con’t) A C B 115° a = 30 c b
30°
a = 30 c b
35°
Set up the Law of Sines to find side b:

12. Example 2 (con’t) A C B 115° a = 30 c b = 26.2
30°
a = 30 c
b = 26.2
35°
Set up the Law of Sines to find side c:

13. Example 2 (solution) Angle A = 35° Side b = 26.2 cm Side c = 47.4 cm A
115°
30°
a = 30
c = 47.4
b = 26.2
35°
Angle A = 35°
Side b = 26.2 cm
Side c = 47.4 cm
Note: Use the Law of Sines whenever you are given 2 angles and one side!

14. The Ambiguous Case (SSA)
When given SSA (two sides and an angle that is NOT the included angle) , the situation is ambiguous. The dimensions may not form a triangle, or there may be 1 or 2 triangles with the given dimensions. We first go through a series of tests to determine how many (if any) solutions exist.

15. The Ambiguous Case (SSA)
In the following examples, the given angle will always be angle A and the given sides will be sides a and b. If you are given a different set of variables, feel free to change them to simulate the steps provided here.
‘a’ - we don’t know what angle C is so we can’t draw side ‘a’ in the right position A
B ? b
C = ?
c = ?

16. The Ambiguous Case (SSA)
Situation I: Angle A is obtuse
If angle A is obtuse there are TWO possibilities
If a ≤ b, then a is too short to reach side c - a triangle with these dimensions is impossible.
If a > b, then there is ONE triangle with these dimensions. A
B ? a b
C = ?
c = ? A
B ? a b
C = ?
c = ?

17. The Ambiguous Case (SSA)
Situation I: Angle A is obtuse - EXAMPLE
Given a triangle with angle A = 120°, side a = 22 cm and side b = 15 cm, find the other dimensions.
Since a > b, these dimensions are possible. To find the missing dimensions, use the Law of Sines: A B
a = 22
15 = b C c
120°

18. The Ambiguous Case (SSA)
Situation I: Angle A is obtuse - EXAMPLE
Angle C = 180° - 120° ° = 23.8°
Use Law of Sines to find side c: A B
a = 22
15 = b C c
120°
36.2°
Solution: angle B = 36.2°, angle C = 23.8°, side c = 10.3 cm

19. The Ambiguous Case (SSA)
Situation II: Angle A is acute
If angle A is acute there are SEVERAL possibilities.
Side ‘a’ may or may not be long enough to reach side ‘c’. We calculate the height of the altitude from angle C to side c to compare it with side a. A
B ? b
C = ?
c = ? a

20. The Ambiguous Case (SSA)
Situation II: Angle A is acute
First, use SOH-CAH-TOA to find h: A
B ? b
C = ?
c = ? a h
Then, compare ‘h’ to sides a and b . . .

21. The Ambiguous Case (SSA)
Situation II: Angle A is acute
If a < h, then NO triangle exists with these dimensions. A
B ? b
C = ?
c = ? a h

22. The Ambiguous Case (SSA)
Situation II: Angle A is acute
If h < a < b, then TWO triangles exist with these dimensions. A B b C c a h A B b C c a h
If we open side ‘a’ to the outside of h, angle B is acute.
If we open side ‘a’ to the inside of h, angle B is obtuse.

23. The Ambiguous Case (SSA)
Situation II: Angle A is acute
If h < b < a, then ONE triangle exists with these dimensions.
Since side a is greater than side b, side a cannot open to the inside of h, it can only open to the outside, so there is only 1 triangle possible! A B b C c a h

24. The Ambiguous Case (SSA)
Situation II: Angle A is acute
If h = a, then ONE triangle exists with these dimensions. A B b C c
a = h
If a = h, then angle B must be a right angle and there is only one possible triangle with these dimensions.

25. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
Given a triangle with angle A = 40°, side a = 12 cm and side b = 15 cm, find the other dimensions.
Find the height: A
B ?
15 = b
C = ?
c = ?
a = 12 h
40°
Since a > h, but a< b, there are 2 solutions and we must find BOTH.

26. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
FIRST SOLUTION: Angle B is acute - this is the solution you get when you use the Law of Sines! A B
15 = b C c
a = 12 h
40°

27. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
SECOND SOLUTION: Angle B is obtuse - use the first solution to find this solution.
In the second set of possible dimensions, angle B is obtuse, because side ‘a’ is the same in both solutions, the acute solution for angle B & the obtuse solution for angle B are supplementary.
Angle B = ° = 126.5°
a = 12 A B
15 = b C c
40°
1st ‘a’
1st ‘B’

28. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 1
SECOND SOLUTION: Angle B is obtuse
Angle B = 126.5°
Angle C = 180°- 40° ° = 13.5°
a = 12 A B
15 = b C c
40°
126.5°

29. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EX. 1 (Summary)
Angle B = 53.5°
Angle C = 86.5°
Side c = 18.6
Angle B = 126.5°
Angle C = 13.5°
Side c = 4.4
a = 12 A B
15 = b C
c = 4.4
40°
126.5°
13.5° A B
15 = b C
c = 18.6
a = 12
40°
53.5°
86.5°

30. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 2
Given a triangle with angle A = 40°, side a = 12 cm and side b = 10 cm, find the other dimensions. A
B ?
10 = b
C = ?
c = ?
a = 12 h
40°
Since a > b, and h is less than a, we know this triangle has just ONE possible solution - side ‘a’opens to the outside of h.

31. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE 2
Using the Law of Sines will give us the ONE possible solution: A B
10 = b C c
a = 12
40°

32. The Ambiguous Case - Summary
if angle A is acute
find the height,
h = b*sinA
if angle A is obtuse
if a < b  no solution
if a > b  one solution
if a < h  no solution
if h < a < b  2 solutions
one with angle B acute,
one with angle B obtuse
if a > b > h  1 solution
If a = h  1 solution
angle B is right
(Ex I)
(Ex II-1)
(Ex II-2)

33. The Law of Sines AAS ASA SSA (the ambiguous case)
Use the Law of Sines to find the missing dimensions of a triangle when given any combination of these dimensions.

34. Area of an Oblique Triangle

35. 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.

36. Area of a Triangle - SAS B a h C Looking at this from all three sides:
SAS – you know two sides: b, c and the angle between: A
Remember area of a triangle is ½ base ● height
Base = b
Height = c ● sin A
 Area = ½ bc(sinA) A B C c a b h
Looking at this from all three sides:
Area = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

37. Area of an Oblique Triangle

38. 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 = ½ ab sin C
= ½ (90)(52)(sin102)
 2289 square meters.