1. Warm Up Chapter 3 Solve and graph on a number line: −𝟑𝒙−𝟗 ≥𝟗
Solve on a number line: 𝟐𝒙−𝟓 𝟐 (𝒙+𝟑)(𝒙+𝟐)>𝟎
A factory produces short and long sleeved shirts. A short sleeved shirt requires 30 minutes of labor, a long sleeved shirt requires 45 minutes of labor. There are 240 hours of labor available each day. The maximum number of shirts that can be packaged is If the profit on a short sleeved shirt is $11 and a long sleeved shirt is $16, find the maximum daily profit.
*Write the constraints, profit equation, and make a graph

2. Warm Up Chapter 3 −3𝑥−9≥9 −3𝑥−9≤−9 −3𝑥≤0 −3𝑥≥18 𝑥≥0 𝑥≤−6
Solve and graph on a number line:
−𝟑𝒙−𝟗 ≥𝟗
−3𝑥−9≥9
−3𝑥−9≤−9
−3𝑥≥18
−3𝑥≤0
𝑥≥0
𝑥≤−6

3. Warm Up Chapter 3
2. Solve on a number line: 𝟐𝒙−𝟓 𝟐 (𝒙+𝟑)(𝒙+𝟐)>𝟎

4. Warm Up Chapter 3 2. Solve on a number line: 𝟐𝒙−𝟓 𝟐 (𝒙+𝟑)(𝒙+𝟐)>𝟎
𝑥<−3
𝑥>−2
𝑥≠2.5

5. Warm Up Chapter 3 𝒔≥𝟎 𝟏 𝟐 𝒔+ 𝟑 𝟒 𝒈≤𝟐𝟒𝟎 𝒈≥𝟎 𝒔+𝒈≤𝟒𝟎𝟎 𝑷=𝟏𝟏𝒔+𝟏𝟔𝒈
3. A factory produces short and long sleeved shirts. A short sleeved shirt requires 30 minutes of labor, a long sleeved shirt requires 45 minutes of labor. There are 240 hours of labor available each day. The maximum number of shirts that can be packaged is 400.
𝒔≥𝟎
𝟏 𝟐 𝒔+ 𝟑 𝟒 𝒈≤𝟐𝟒𝟎
𝒈≥𝟎
Let s = short sleeved shirts
Let g = long sleeved shirts
𝒔+𝒈≤𝟒𝟎𝟎
If the profit on a short sleeved shirt is $11 and a long sleeved shirt is $16, find the maximum daily profit.
𝑷=𝟏𝟏𝒔+𝟏𝟔𝒈

6. The maximum daily profit is $5200𝒈
𝑷=𝟏𝟏𝒔+𝟏𝟔𝒈
𝒔≥𝟎
𝒔+𝒈≤𝟒𝟎𝟎
450
(0,320)
P= =$5120
𝒈≥𝟎
400
𝟏 𝟐 𝒔+ 𝟑 𝟒 𝒈≤𝟐𝟒𝟎
350
(0,320)
(240,160)
P= =$5200
300
250
200
(240,160)
(400,0)
P= =$4400
150
100
The maximum daily profit is $5200 50
(0,400) 𝒔 50
100
150
200
250
300
350
400
450

7. Objective: To use the law of sines to find unknown parts of a triangle
Section Law of Sines
Objective: To use the law of sines to find unknown parts of a triangle

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

9. 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.

10. Example 1 The angles in a ∆ total 180°, so angle C = 30°.
Set up the Law of Sines to find side b: A C B
70°
80°
a = 12 c b

11. Set up the Law of Sines to find side c:
Example 1
Set up the Law of Sines to find side c: A C B
70°
80°
a = 12 c
b = 12.6
30°

12. Angle C = 30° Side b = 12.6 cm Side c = 6.4 cm Example 1 A C B 70° 80°

13. 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.

14. Example 2
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

15. Example 2 Set up the Law of Sines to find side b: A C B 115° a = 30 c
30°
a = 30 c b
35°

16. Set up the Law of Sines to find side c:
Example 2
Set up the Law of Sines to find side c: A C B
115°
30°
a = 30 c
b = 26.2
35°

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

18. 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.

19. The Ambiguous Case (SSA)
if the given angle is acute
find the height,
h = adj ∙sinA
if the given angle is obtuse
if opp < adj  no solution
if opp > adj  one solution
if opp < h  no solution
if h < opp < adj  2 solutions
The two solutions are supplementary.
One angle B is acute, one angle B is obtuse
if opp > adj  1 solution
If opp = h  1 solution
angle B is right B A
adj
opp h
𝐴= given angle

20. 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.
‘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 = ?

21. 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 = ?

22. 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

23. 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 . . .

24. 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

25. 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.

26. The Ambiguous Case (SSA)
Situation II: Angle A is acute
If a > b, 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

27. 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.

28. one solution How many solutions? opp=15, adj= 10 Acute or obtuse?

29. no solution How many solutions? opp=10, adj= 35 Acute or obtuse?

30. one solution How many solutions? opp=8, adj= 3 Acute or obtuse? acute

31. no solution How many solutions? opp=5, adj= 7 opp < h
Acute or obtuse?
acute
opp=5, adj= 7
opp < adj
Find h:
h = adj ∙sinA
= 7 ∙sin62
= 6.2
no solution
opp < h

32. two solutions How many solutions? opp=19, adj= 21 h < opp < adj
Acute or obtuse?
acute
opp=19, adj= 21
opp < adj
Find h:
h = adj ∙sinA
= 21 ∙sin62
= 18.5
two solutions
h < opp < adj

33. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE
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 (adj) C c
a = 12 (opp) h
40°
Since h < opp (a) < adj (b), there are 2 solutions and we must find BOTH.

34. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE
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°

35. The Ambiguous Case (SSA)
Situation II: Angle A is acute - EXAMPLE
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’

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

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

38. Homework
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