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 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

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

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

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

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

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

Objective: To use the law of sines to find unknown parts of a triangle Section 9.3 Law of Sines Objective: To use the law of sines to find unknown parts of a triangle

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

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.

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

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°

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

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.

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

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°

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°

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!

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.

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

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 = ?

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 = ?

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

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 . . .

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

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.

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

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.

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

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

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

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

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

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.

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°

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 = 180 - 53.5° = 126.5° a = 12 A B 15 = b C c 40° 1st ‘a’ 1st ‘B’

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°- 126.5° = 13.5°

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°

Homework Page 347 #1,2,3-21 odds