1. 6.1 Law of Sines Objective To use Law of Sines to solve oblique triangles and to find the areas of oblique triangles.

2. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2 An oblique triangle is a triangle that has no right angles. Definition: Oblique Triangles To solve an oblique triangle, you need to know the measure of at least one side and the measures of any other two parts of the triangle – two sides, two angles, or one angle and one side. C BA a b c

3. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 3 The following cases are considered when solving oblique triangles. Solving Oblique Triangles 1.Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) A C c A B c a c b C c a c a B

4. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 4 The first two cases can be solved using the Law of Sines. (The last two cases can be solved using the Law of Cosines.) Definition: Law of Sines Law of Sines If ABC is an oblique triangle with sides a, b, and c, then Acute Triangle C BA b h c a C B A b h c a Obtuse Triangle

5. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 5 Find the remaining angle and sides of the triangle. Example: Law of Sines - ASA Example 1 (ASA): The third angle in the triangle is A = 180  – A – B = 180  – 10  – 60  = 110 . C B A b c 60  10  a = 4.5 ft 110  Use the Law of Sines to find side b and c. 4.15 ft 0.83 ft

6. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 6 Use the Law of Sines to solve the triangle. A = 110 , a = 125 inches, b = 100 inches Example: Single Solution Case - SSA Example 2 (SSA): C  180  – 110  – 48.74  C B A b = 100 in c a = 125 in 110  48.74  21.26  48.23 in = 21.26 

7. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 7 The Ambiguous Case (SSA) Two angles and one side determine a unique triangle. But what if you are given two sides and one opposite angle? Three possible situations can occur: 1) no such triangle exists; 2) one such triangle exits; 3) two distinct triangles may satisfy the conditions.

8. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 8 The Ambiguous Case (SSA A is acute a < b a > b * sin A Two Solutions a = b * sin A One Solution a < b * sin A No Solutions

9. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 9 The Ambiguous Case (SSA) A is acute and a ≥ b One Solution

10. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 10 The Ambiguous Case (SSA) A is obtuse a > b One Solution a < b No Solution

11. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 11 Use the Law of Sines to solve the triangle. A = 76 , a = 18 inches, b = 20 inches Example: No-Solution Case - SSA Example 3 (SSA): There is no angle whose sine is 1.078. There is no triangle satisfying the given conditions. C A B b = 20 in a = 18 in 76 

12. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 12 Use the Law of Sines to solve the triangle. A = 58 , a = 11.4 cm, b = 12.8 cm Example: Two-Solution Case - SSA Example 4 (SSA): 72.2  10.3 cm Two different triangles can be formed. 49.8  a = 11.4 cm C A B1B1 b = 12.8 cm c 58  Example continues. C  180  – 58  – 72.2  = 49.8 

13. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 13 Use the Law of Sines to solve the second triangle. A = 58 , a = 11.4 cm, b = 12.8 cm Example: Two-Solution Case – SSA continued Example (SSA) continued: B 2  180  – 72.2  = 107.8  107.8  C A B2B2 b = 12.8 cm c a = 11.4 cm 58  14.2  3.3 cm 72.2  10.3 cm 49.8  a = 11.4 cm C A B1B1 b = 12.8 cm c 58  C  180  – 58  – 107.8  = 14.2 

14. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 14 Area of an Oblique Triangle C BA b c a Find the area of the triangle. A = 74 , b = 103 inches, c = 58 inches Example 5: 74  103 in 58 in

15. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 15 Example 6 Finding the Area of a Triangular Lot Find the area of a triangular lot containing side lengths that measure 24 yards and 18 yards and form an angle of 80° A = ½(18)(24)sin80 A = 212.7 yards

16. Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 16 Application A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 14  with the horizontal. The flagpole casts a 16-meter shadow up the slope when the angle of elevation from the tip of the shadow to the sun is 20 . How tall is the flagpole? Example 7 Application: 20  Flagpole height: b 70  34  16 m 14  The flagpole is approximately 9.5 meters tall. B A C