2. Copyright © 2007 Pearson Education, Inc. Slide 10-2 Chapter 10: Applications of Trigonometry; Vectors 10.1The Law of Sines 10.2The Law of Cosines and Area Formulas 10.3Vectors and Their Applications 10.4Trigonometric (Polar) Form of Complex Numbers 10.5Powers and Roots of Complex Numbers 10.6Polar Equations and Graphs 10.7More Parametric Equations

3. Copyright © 2007 Pearson Education, Inc. Slide 10-3 10.1The Law of Sines Congruence Axioms Side-Angle-Side (SAS)If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent. Angle-Side-Angle (ASA)If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent. Side-Side-Side (SSS)If three sides of one triangle are equal to three sides of a second triangle, the triangles are congruent.

4. Copyright © 2007 Pearson Education, Inc. Slide 10-4 Recall In a triangle, the sum of the interior angles is 180º. No triangle can have two obtuse angles. The height of a triangle is less than or equal to the length of two of the sides. The sine function has a range of If the θ is a positive decimal < 1, the θ can lie in the first quadrant (acute <) or in the second quadrant (obtuse <)..

5. Copyright © 2007 Pearson Education, Inc. Slide 10-5 10.1Data Required for Solving Oblique Triangles Case 1One side and two angles known: »SAA or ASA Case 2Two sides and one angle not included between the sides known: »SSA »This case may lead to more than one solution. Case 3Two sides and one angle included between the sides known: »SAS Case 4Three sides are known: »SSS

6. Copyright © 2007 Pearson Education, Inc. Slide 10-6 10.1Derivation of the Law of Sines Start with an acute or obtuse triangle and construct the perpendicular from B to side AC. Let h be the height of this perpendicular. Then c and a are the hypotenuses of right triangle ADB and BDC, respectively.

7. Copyright © 2007 Pearson Education, Inc. Slide 10-7 10.1The Law of Sines In a similar way, by constructing perpendiculars from other vertices, the following theorem can be proven. Law of Sines In any triangle ABC, with sides a, b, and c, Alternative forms are sometimes convenient to use:

8. Copyright © 2007 Pearson Education, Inc. Slide 10-8 10.1Using the Law of Sines to Solve a Triangle ExampleSolve triangle ABC if A = 32.0°, B = 81.8°, and a = 42.9 centimeters. SolutionDraw the triangle and label the known values. Because A, B, and a are known, we can apply the law of sines involving these variables.

9. Copyright © 2007 Pearson Education, Inc. Slide 10-9 10.1Using the Law of Sines to Solve a Triangle To find C, use the fact that there are 180° in a triangle. Now we can find c

10. Copyright © 2007 Pearson Education, Inc. Slide 10-10 10.1Using the Law of Sines in an Application (ASA) ExampleTwo stations are on an east-west line 110 miles apart. A forest fire is located on a bearing of N 42° E from the western station at A and a bearing of N 15° E from the eastern station at B. How far is the fire from the western station? SolutionAngle BAC = 90° – 42° = 48° Angle B = 90° + 15° = 105° Angle C = 180° – 105° – 48° = 27° Using the law of sines to find b gives

11. Copyright © 2007 Pearson Education, Inc. Slide 10-11 10.1Ambiguous Case If given the lengths of two sides and the angle opposite one of them, it is possible that 0, 1, or 2 such triangles exist. Some basic facts that should be kept in mind: –For any angle , –1  sin   1, if sin  = 1, then  = 90° and the triangle is a right triangle. –sin  = sin(180° –  ). –The smallest angle is opposite the shortest side, the largest angle is opposite the longest side, and the middle-value angle is opposite the intermediate side (assuming unequal sides).

12. Copyright © 2007 Pearson Education, Inc. Slide 10-12 10.1Number of Triangles Satisfying the Ambiguous Case Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate sin B.) 1.If sin B > 1, then no triangle satisfies the given conditions. 2.If sin B = 1, then one triangle satisfies the given conditions and B = 90°. 3.If 0 < sin B < 1, then either one or two triangles satisfy the given conditions (a)If sin B = k, then let B 1 = sin -1 k and use B 1 for B in the first triangle. b)Let B 2 = 180° – B 1. If A + B 2 < 180°, then a second triangle exists. In this case, use B 2 for B in the second triangle.

13. Copyright © 2007 Pearson Education, Inc. Slide 10-13 10.1 Ambiguous Case a < b sinA a = b sinA a < b sinA a > b sinA

14. Copyright © 2007 Pearson Education, Inc. Slide 10-14 How does it work? Web demo

15. Copyright © 2007 Pearson Education, Inc. Slide 10-15 10.1Ambiguous Case for Obtuse Angle A

16. Copyright © 2007 Pearson Education, Inc. Slide 10-16 10.1Solving the Ambiguous Case: One Triangle Example Solve the triangle ABC, given A = 43.5°, a = 10.7 inches, and c = 7.2 inches. Solution The other possible value for C: C = 180° – 27.6° = 152.4°. Add this to A: 152.4° + 43.5° = 195.9° > 180° Therefore, there can be only one triangle.

17. Copyright © 2007 Pearson Education, Inc. Slide 10-17 10.1Solving the Ambiguous Case: One Triangle

18. Copyright © 2007 Pearson Education, Inc. Slide 10-18 10.1Solving the Ambiguous Case: No Such Triangle ExampleSolve the triangle ABC if B = 55°40´, b = 8.94 meters, and a = 25.1 meters. SolutionUse the law of sines to find A. Since sin A cannot be greater than 1, the triangle does not exist.

19. Copyright © 2007 Pearson Education, Inc. Slide 10-19 Example Solve the triangle ABC if A = 55.3°, a = 22.8 feet, and b = 24.9 feet. Solution 10.1Solving the Ambiguous Case: Two Triangles

20. Copyright © 2007 Pearson Education, Inc. Slide 10-20 10.1Solving the Ambiguous Case: Two Triangles To see if B 2 = 116.1° is a valid possibility, add 116.1° to the measure of A: 116.1° + 55.3° = 171.4°. Since this sum is less than 180°, it is a valid triangle. Now separate the triangles into two: AB 1 C 1 and AB 2 C 2.

21. Copyright © 2007 Pearson Education, Inc. Slide 10-21 10.1Solving the Ambiguous Case: Two Triangles Now solve for triangle AB 2 C 2.

22. Copyright © 2007 Pearson Education, Inc. Slide 10-22 Practice: Answer in pairs. Find m  B, m  C, and c, if they exist. 1) a = 9.1, b = 12, m  A = 35 o 2) a = 25, b = 46, m  A = 37 o 3) a = 15, b = 10, m  A = 66 o

23. Copyright © 2007 Pearson Education, Inc. Slide 10-23 Answers: 1)Case 1: m  B=49 o,m  C=96 o,c=15.78 Case 2: m  B=131 o,m  C=14 o,c=3.84 2)No possible solution. 3)m  B=38 o,m  C=76 o,c=15.93

24. Copyright © 2007 Pearson Education, Inc. Slide 10-24 Homework P. 436 #1-5 all #8-18, evens DUE FRIDAY – we will grade it 5 minutes after the tardy bell rings