1. Solve and graph on a number line: 3x 9 9
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1 1. Solve and graph on a number line: 3x Solve on a number line: 2x 5 2 (x + 3)(x + 2) > 0 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. 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 1. Solve and graph on a number line: 3x 9 9 3x 9 9 3x 18 x 6 3x 9 9 3x 0 x 0
3 2. Solve on a number line: 2x 5 2 (x + 3)(x + 2) > 0
4 2. Solve on a number line: 2x 5 2 (x + 3)(x + 2) > 0 x < 3 x > 2 x 2.5
5 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. Let s = short sleeved shirts Let g = long sleeved shirts s 0 g 0 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 1 2 s s + g 400 g 240
6 450 g s 0 g (0,320) s + g s g 240 P = 11s + 16g (0,320) P = = $5120 (240,160) P = = $ (240,160) (400,0) P = = $ (0,400) s The maximum daily profit is $5200
7 Section 9.3 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: a sin A b sin B c sin C
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, B so angle C = 30. c 80 a = 12 Set up the Law of Sines to find side b: A 70 b C 12 sin 70 b sin sin 80 b sin 70 b 12 sin80 sin cm
11 Example 1 Set up the Law of B Sines to find side c: c 80 a = sin 70 c sin 30 A 70 b = C 12 sin 30 c sin70 c 12 sin 30 sin70 6.4cm
12 Example 1 B 80 a = 12 Angle C = 30 Side b = 12.6 cm A 70 b = C Side c = 6.4 cm
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 B To solve for the missing sides or angles, we must have an angle and 30 c opposite side to set up the first equation. a = 30 We MUST find angle A first because C 115 b A the only side given is side a. The angles in a total 180, so angle A = 35.
15 Example 2 Set up the Law of B Sines to find side b: 30 a = 30 C c 115 b 35 A 30 sin35 b sin sin 30 b sin35 b 30 sin30 sin cm
16 Example 2 Set up the Law of B Sines to find side c: a = c 30 sin35 c sin sin115 c sin35 C b = 26.2 A c 30 sin115 sin cm
17 Example 2 B Angle A = Side b = 26.2 cm c = 47.4 Side c = 47.4 cm a = 30 C b = 26.2 A 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 obtuse if the given angle is acute find the height, h = adj sina 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 A adj B h opp 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. b C =? a - we don t know what angle C is so we can t draw side a in the right position A B? c =?
21 The Ambiguous Case (SSA) Situation I: Angle A is obtuse If angle A is obtuse there are TWO possibilities C =? If a b, then a is too short to reach side c - a triangle with these dimensions is impossible. C =? If a > b, then there is ONE triangle with these dimensions. b a b a A B? c =? A B? c =?
22 The Ambiguous Case (SSA) Situation II: Angle A is acute If angle A is acute there are SEVERAL possibilities. b C =? a 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? c =?
23 The Ambiguous Case (SSA) Situation II: Angle A is acute C =? First, use SOH-CAH-TOA to find h: b h A B? c =? a sin A h b h bsin A 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. C =? b a h A c =? B?
25 The Ambiguous Case (SSA) Situation II: Angle A is acute If h < a < b, then TWO triangles exist with these dimensions. C C b h a b a h A c B A c B 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. A b C h c a B 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!
27 The Ambiguous Case (SSA) Situation II: Angle A is acute If h = a, then ONE triangle exists with these dimensions. A b c C a = h B If a = h, then angle B must be a right angle and there is only one possible triangle with these dimensions.
28 How many solutions? Acute or obtuse? obtuse opp=15, adj= 10 opp > adj one solution
29 How many solutions? Acute or obtuse? obtuse opp=10, adj= 35 opp < adj no solution
30 How many solutions? Acute or obtuse? acute opp=8, adj= 3 opp > adj one solution
31 How many solutions? Acute or obtuse? acute opp=5, adj= 7 opp < adj Find h: h = adj sina= 7 sin62 = 6.2 opp < h no solution
32 How many solutions? Acute or obtuse? acute opp=19, adj= 21 opp < adj Find h: h = adj sina= 21 sin62 = 18.5 h < opp < adj two solutions
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. C Find the height: A 15 = b (adj) 40 c h a = 12 (opp) B h bsin A h 15sin 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 15 = b 40 c C h a = 12 B 12 sin sin B B sin 1 15sin C c sin sin 40 c 12sin sin 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, C because side a is the same in both 1st a 15 = b solutions, the acute solution for a = 12 angle B & the obtuse solution for A angle B are supplementary. 40 c B 1st B Angle B = = 126.5
36 The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE SECOND SOLUTION: Angle B is obtuse A 15 = b a = c B C Angle B = Angle C = = 13.5 c sin sin 40 c 12sin sin 40
37 The Ambiguous Case (SSA) Situation II: Angle A is acute - Example Angle B = 53.5 Angle C = 86.5 Side c = 18.6 C Angle B = Angle C = 13.5 Side c = C 15 = b 86.5 a = = b a = 12 A c = 18.6 B A c = 4.4 B
38 Homework Page 347 #1,2,3-21 odds
2. Factor and find all the zeros: b. p 6 + 7p 3 30 = Identify the domain: 4. Simplify:
1. Divide: 5x 5 3x 3 + 2x 2 8x + 1 by x + 3 2. Fator and find all the zeros: a. x 3 + 5x 2 3x 15 = 0 b. p 6 + 7p 3 30 = 0 3. Identify the domain: a. f x = 3x 5x 2x 15 4. Simplify: a. 3x2 +6x+3 3x+3 b.
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