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Warm Up Chapter 4 1. 2. 3. Find the inverse of the function 3 π(π₯) = 2 + π₯ β 5 Sketch the graph of g and g-1.Then find a rule for g-1(x). 2 π π₯ = π₯ + 2, π₯ β₯ 0 A manufacturer wants to design an open top box with a square base and a surface area of 400 in2. a. Express the volume V of the box as a function of the base width w. b. What dimensions will produce a box with maximum volume? Warm Up Chapter 4 1. 3 Find the inverse of the function π(π₯) = 2 + π₯ β 5, π¦ =2+ π₯ =2+ π₯β2= π₯β2 π₯β2 3 3 3 3 3 π₯β5 π¦β5 π¦β5 =π¦β5 +5=π¦ π β1 3 (π₯) = π₯ β 2 + 5 Warm Up Chapter 4 2. Sketch the graph of g and g-1. π π₯ = π₯ 2 + 2, π₯β₯0 Then find a rule for g-1(x). 2 g π₯ =π¦ +2 π₯ β 2 = π¦2 g π₯β2=π¦ g-1 π β1 π₯ = π₯ β 2, π₯β₯2 Warm Up Chapter 4 3. A manufacturer wants to design an open top box with a square base and a surface area of 400 in2. a. Express the volume V of the box as a function of the base width w. V (w, h) ο½ w h 2 ππ΄: π€ 2 + 4π€β = 400 4π€β = 400 βπ€ 2 400 βπ€ 2 β= 4π€ h Write h in terms of w. π π€ = π€2 w w 400 βπ€ 2 4π€ 400π€ βπ€ π π€ = 4 3 Warm Up Chapter 4 3. A manufacturer wants to design an open top box with a square base and a surface area of 400 in2. b. What dimensions will produce a box with maximum volume? 400π€ βπ€ π π€ = 4 w = 11.55 ππ 3 Calculate max value on the calculator. (11.55,769.80) h w w 400 βπ€ 2 400 β 11.552 β= = = 5.77 ππ 4π€ 4(11.55) 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: a b c ο½ ο½ sin A sin B sin C Use Law of SINES when you are given... ο΅ AAS - 2 angles and 1 adjacent side ο΅ ASA - 2 angles and their included side ο΅ SSA - 2 sides and 1 angle that is not the included angle (ambiguous case) 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 β total 180°, so angle C = 30°. B 80° a = 12 c A 70° Set up the Law of Sines to find side b: b C 12 b ο½ sin 70ο° sin 80ο° 12 οsin 80ο° ο½ b οsin 70ο° 12 οsin 80ο° bο½ ο» 12.6cm sin 70ο° Example 1 Set up the Law of Sines to find side c: B 80° c A 70° a = 12 b = 12.6 30° 12 c ο½ sin 70ο° sin 30ο° C 12 οsin 30ο° ο½ c ο sin 70ο° 12 οsin 30ο° cο½ ο» 6.4cm sin 70ο° Example 1 B 80° A 70° Angle C = 30° a = 12 b = 12.6 30° Side b = 12.6 cm C Side c = 6.4 cm 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. B 30° c We MUST find angle A first because the only side given is side a. a = 30 C 115° b A The angles in a β total 180°, so angle A = 35°. Example 2 Set up the Law of Sines to find side b: B 30 b ο½ sin 35ο° sin 30ο° 30° c a = 30 115° 35° C b 30 οsin 30ο° ο½ b οsin 35ο° A 30 οsin 30ο° bο½ ο» 26.2cm sin 35ο° Example 2 Set up the Law of Sines to find side c: B 30 c ο½ sin 35ο° sin 115ο° 30° c a = 30 30 οsin 115ο° ο½ c ο sin 35ο° C 115° 35° b = 26.2 A 30 οsin 115ο° cο½ ο» 47.4cm sin 35ο° Example 2 Angle A = 35° B Side b = 26.2 cm 30° c = 47.4 Side c = 47.4 cm a = 30 C 115° 35° b = 26.2 A 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 if opp < adj ο no solution if opp > adj ο one solution obtuse if opp < h ο no solution adj if the given angle is if h < opp < adj ο 2 solutions A one angle is B acute, one angle B is acute obtuse find the height, h = adj βsinA (supplementary) if opp > adj ο 1 solution If opp = h ο 1 solution angle B is right opp h π΄ = given angle 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: C 15 = b (adj) A 40° h c a = 12 (opp) B h ο½ bsin A h ο½ 15sin 40ο° ο½ 9.6 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! C a = 12 15 = b A h 40° c B 12 15 ο½ sin 40ο° sin B 15sin 40ο° οΆο B ο½ sin ο1 ο¦ο ο½ 53.5ο° ο¨ο 12 οΈο C ο½ 180ο° ο 40ο° ο 53.5ο° ο½ 86.5ο° c 12 ο½ sin 86.5ο° sin 40ο° 12sin 86.5ο° cο½ ο½ 18.6 sin 40ο° 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 βaβ is the same in both 1st βaβ solutions, the acute solution for 15 = b a = 12 angle B & the obtuse solution for 40° A angle B are supplementary. c B 1st βBβ Angle B = 180 - 53.5° = 126.5° The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE SECOND SOLUTION: Angle B is obtuse C Angle C = 180°- 40°- 126.5° = 13.5° 15 = b A a = 12 40° 126.5° c Angle B = 126.5° B c 12 ο½ sin13.5ο° sin 40ο° 12sin 13.5ο° cο½ ο½ 4.4 sin 40ο° 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 C 15 = b A 40° 86.5° 13.5° 15 = b a = 12 53.5° B c = 18.6 A 40° 126.5° c = 4.4 B C a = 12 Homework Page 347 #1,2,3-21 odds