1. Math 112 Elementary Functions Section 1 The Law of Sines Chapter 7 – Applications of Trigonometry

2. Solving Right Triangles – Revisited! Solving Triangles? Using given information, determine the lengths of the sides and measures of the angles. What must be known to solve a right triangle? Lengths of two sides. Length of a side and the measure of an acute angle. How do you solve the triangle? sin  = opp/hyp cos  = adj/hyp tan  = opp/adj sin  = opp/hyp cos  = adj/hyp tan  = opp/adj

3. Solving Oblique Triangles – Five Cases Given 1 side and 2 angles AAS and ASA 60° 40° 20 60° 40° 25

4. Solving Oblique Triangles – Five Cases Given 1 side and 2 angles AAS and ASA Given 2 sides and 1 angle SSA and SAS 60° 20 25 40° 25 20

5. Solving Oblique Triangles – Five Cases Given 1 side and 2 angles AAS and ASA Given 2 sides and 1 angle SSA and SAS Given 3 sides SSS 20 25 13

6. Solving Oblique Triangles – Five Cases Law of Sines (this section) Used to solve AAS, ASA, and SSA triangles. Law of Cosines (next section) Used to solve SAS and SSS triangles. 60° 40° 20 60° 40° 25 60° 20 25 40° 25 20 25 13

7. a b c A B C The Law of Sines – Acute Triangle h

8. The Law of Sines – Obtuse Triangle a b c A B C h

9. The Law of Sines a b c A B C

10. Solving Oblique Triangles – AAS w/ the Law of Sines 1. Find the third angle.  = 180° - (60° + 40°) = 80° 2. Use the law of sines to find a second side. x/sin 40° = 20/sin 60° x  14.8 3. Use the law of sines to find the third side. y/sin 80° = 20/sin 60° y  22.7 60° 40° 20 x y  NOTE: Always use EXACT values if possible.

11. Solving Oblique Triangles – ASA w/ the Law of Sines 60° 40° 25 1. Find the third angle.  = 180° - (60° + 40°) = 80° 2. Use the law of sines to find a second side. x/sin 40° = 25/sin 80° x  16.3 3. Use the law of sines to find the third side. y/sin 60° = 25/sin 80° y  22.0 x y  NOTE: Always use EXACT values if possible.

12. Solving Oblique Triangles – SSA w/ the Law of Sines With AAS and ASA, the given data will determine a unique triangle. With SSA, the given data could determine … no triangle one triangle two triangles 60° 20 25 ?

13. Solving Oblique Triangles – SSA w/ the Law of Sines Case 1: No Solution The side opposite the given angle is not long enough to reach the other side of the angle. 40° 22 8 A C B ?

14. Solving Oblique Triangles – SSA w/ the Law of Sines Case 2a: One Solution The side opposite the given angle is just barely long enough to reach the other side of the angle. A C B ? 30° 22 11 NOTE: Angle C and side c still need to be determined.

15. Solving Oblique Triangles – SSA w/ the Law of Sines Case 3: Two Solutions The side opposite the given angle is more than long enough to reach the other side of the angle but is shorter than the other given side. 40° 22 20 A C B ? 20 NOTE: Angle C and side c still need to be determined for BOTH solutions.

16. Solving Oblique Triangles – SSA w/ the Law of Sines Case 2b: One Solution The side opposite the given angle is more than long enough to reach the other side of the angle but is longer than the other given side. B ? 40° 22 25 A C NOTE: Angle C and side c still need to be determined. Since 145.6 + 40  180, this solution is invalid.

17. The Area of a Triangle C b a h area = ½bh sin C = h / a  h = a sin C Therefore, … area = ½ ab sin C In general, the area of a triangle is half the product of two sides times the sine of the included angle.