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P.o.D. – Find the amplitude, period, and vertical translation of each graph. 1.) 𝑦 = 3 sin(2𝑥 ) − 5 2.) 𝑦 = cos(𝑥) 3.) 𝑦 = −4 sin(0.5𝑥 ) + 2 4.) 𝑦 = −2 cos(3𝑥 ) − 6 1 1 3 3 5.) 𝑦 = sin ( 𝑥) + 6.) 𝑦 = − cos(6𝑥 ) − 7 7.) 𝑦 = sin(𝑥 ) + 𝜋 1.) 2.) 3.) 4.) A=3, P=180, VT=-5 A=1, P=360, VT=0 A=4, P=720, VT=2 A=2, P=120, VT=-6 5.) A=1, P=1080, VT=1/3 6.) A=1, P=60, VT=-7 7.) A=1, P=360, VT=pi 10-7: The Law of Sines Learning Target(s): I can determine the measure of an angle given its sine, cosine, or tangent; find the missing sides and angles of a triangle using the Law of Sines; identify and use theorems relating sines and cosines; solve real-world problems using the Law of Sines. Essential Question: How do you use trigonometry to solve and find the areas of oblique triangles? Oblique Triangle = a triangle that has no right angle; could be acute or obtuse. Law of Sines: C b A a c B 𝑎 𝑏 𝑐 = = sin 𝐴 sin 𝐵 sin 𝐶 EX: For triangle ABC, A=30 degrees, B=45 degrees, and a=32 feet. Find the remaining sides and angle. Always begin by drawing a picture. B 45 32 30 A C The easiest thing to find is the missing angle. ∠𝐶 = 180° − 30° − 45° = 105° Now, use the given information to find side b. 𝑏 32 = → sin 45° sin 30° 32 sin 45° 𝑏= = sin 30° 32√2 ≈ 45.25 Next, find side c. 𝑐 32 = → sin 105° sin 30° 32 sin 105° 𝑐= = sin 30° 61.82 Finally, write your solution with all 6 parts of the triangle together. A=30 deg. B=45 deg. C=105 deg. a=32 b=45.25 c=61.82 EX: Because of prevailing winds, a tree grew so that it was leaning 6 degrees from the vertical. At a point 30 meters from the tree, the angle of elevation to the top of the tree is 22.5 degrees. Find the height of the tree. h 96 22.5 30 Begin by finding the missing angle. 180-22.5-96=61.5 degrees Set up the Law of Sine to find h. ℎ 30 = →ℎ sin 22.5° sin 61.5° 30 sin 22.5° = = sin 61.5° 13.06 𝑚𝑒𝑡𝑒𝑟𝑠 EX: For the triangle ABC, a=12inches, b=5inches, and A=31 degrees. Find the remaining sides and angles. We need to find angle B. sin 𝐵 sin 31 = 5 12 5 sin 31 → sin 𝐵 = →𝐵 12 5 sin 31 −1 = 𝑠𝑖𝑛 ( ) = 𝑠𝑖𝑛−1 (.215) 12 = 12.39° Next, find angle C by subtraction. C=180-12.39-31=136.61 degrees. Now, find side c using the Law of Sines. 𝑐 12 = →𝑐 sin 136.61° sin 31° 12 sin 136.61° = = 16.01 sin 31° Write your final answer. A=31 deg. B=12.39 C=136.61 deg deg. a=12 b=5 c=16.01 The Ambiguous Case of the Law of Sines (SSA): Why couldn’t you prove triangle congruency with SSA in geometry? (Draw a picture on the whiteboard) Three possibilities will exist: 1 solution (right triangle), 2 solutions (one acute, one obtuse), or no triangle exists. Case I: If A is an acute angle: If 𝑎 < 𝑏 sin 𝐴, then no triangle exists. If 𝑎 = 𝑏 sin 𝐴, then exactly one right triangle exists. If 𝑎 > 𝑏 sin 𝐴, then two triangles exist. If 𝑎 ≥ 𝑏, then one triangle exists. Case II: If A is an obtuse angle: If a < b, then no triangle exists. If a > b, then one triangle exists. EX: Show that there is no triangle for which A=60degrees, a=4, and b=14. We are in Case I since A is acute. √3 𝑏 sin 𝐴 = 14 sin 60° = 14 ( ) = 7√3 2 ≈ 12.124 In Case I, if a<b sin A, then we have no solution. In this example, 4<12.124. EX: Find two triangles for which A=58degrees, a=4.5, and b=5. Again, we are in Case I since A is acute. We now need to find b sin A. 𝑏 sin 𝐴 = 5 sin 58° ≈ 4.24 Since a>b sin A {4.5>4.24}, we know that we have two solutions. Begin by solving using the Law of Sines. Triangle #1: sin 𝐵 sin 58° 5 sin 58° = → sin 𝐵 = 5 4.5 4.5 5 sin 58° −1 → 𝐵 = 𝑠𝑖𝑛 ( ) 4.5 ≈ 70.437° Use subtraction to find angle C. 𝐶 = 180 − 70.437 − 58 = 51.563° Now, use the Law of Sines to find side c. 𝑐 4.5 = →𝑐 sin 51.563° sin 58° 4.5 sin 51.563° = ≈ 4.156 sin 58° Therefore, the first triangle consists of: A=58 deg. B=70.437 deg. a=4.5 b=5 Triangle #2: C=51.563 deg. c=4.156 ∠𝐵2 = 180 − 𝐵1 = 180 − 70.437 = 109.563° In other words, the second angle B is the supplement of the first angle B. Find angle 𝐶2 by subtraction. 𝐶2 = 180 − 109.563 − 58 = 12.437° Lastly, find side 𝑐2 using the Law of Sines. 𝑐2 4.5 = → 𝑐2 sin 12.437° sin 58° 4.5 sin 12.437° = ≈ sin 58° 1.143 Triangle #2 consists of: 𝐴 = 58° 𝐵2 𝐶2 = 109.563° = 12.437° 𝑎 = 4.5 𝑏=5 𝑐2 = 1.143 Area of a Triangle: The area of any triangle is onehalf the product of the lengths of two sides times the sine of their included angle. That is, 1 1 𝐴𝑟𝑒𝑎 = 𝑎𝑏 sin 𝐶 = 𝑏𝑐 sin 𝐴 2 2 1 = 𝑎𝑐 sin 𝐵 2 EX: Find the area of a triangular lot containing the side lengths that measure 24 yards and 18 yards and form an angle of 80 degrees. 1 𝐴 = (24)(18) sin 80° 2 ≈ 212.718 𝑠𝑞𝑢𝑎𝑟𝑒 𝑦𝑎𝑟𝑑𝑠. *On your own, write a calculator program titled SASAREA that will compute the area of a triangle given two sides and the included angle. EX: On a small lake, a child swam from point A to point B at a bearing of 𝑁 28° 𝐸. The child then swam to point C at a bearing of 𝑁 58° 𝑊. Point C is 800 meters due north of point A. How many total meters did the child swim? (draw a picture and show the work on the whiteboard) About 1056 meters. EX: Suppose you are given a triangle with two angles of 42 degrees and 74 degrees and an included side of 18-m. Find the length of the side opposite the 42 degrees. (draw a triangle on the whiteboard and have a student come to the board to solve) About 13.4m EX: In triangle TRI, angle R is 25 degrees, angle I is 72 degrees, and side r is 13. Find side t. (draw a triangle on the board and have a student come to the board to solve) t=30.53 Upon completion of this lesson, you should be able to: 1. Solve triangles using the Law of Sines. 2. Find the area of a triangle given two side and the included angle. For more information, visit http://www.mathsisfun.com/algebra/trigsine-law.html HW Pg.703 2, 7-12, 15-18 Worksheet 10.7B