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6.1 Law of Sines

6.1 Law of Sines. Objectives:. To use the Law of Sines to 1) solve oblique triangles 2) find the areas of oblique triangles 3) solve applied real-life problems. C. a. b. A. B. c. An oblique triangle is a triangle that has no right angles.

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6.1 Law of Sines

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  1. 6.1 Law of Sines Objectives: To use the Law of Sines to 1) solve oblique triangles 2) find the areas of oblique triangles 3) solve applied real-life problems

  2. C a b A B c An oblique triangle is a triangle that has no right angles. 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. Need to know 3 measurements.

  3. A A c c B C c C a c b c a B a The following cases are considered when solving oblique triangles. • 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)

  4. If ABC is an oblique triangle with sides a, b, and c, then C C a a b h h b A B B c c A The first two cases can be solved using the Law of Sines. (The last two cases can be solved using the Law of Cosines.) Law of Sines Acute Triangle Obtuse Triangle

  5. C 10 a = 4.5 ft b 60 110 B c A Case 1 (ASA): Find the remaining angle and sides of the triangle. The third angle in the triangle is A = 180 – A – B = 180 – 10 – 60 = 110. 4.15 ft 0.83 ft Use the Law of Sines to find side b and c.

  6. C 21.26 a = 125 in b = 100 in 48.74 110 B c A Case 2 (SSA): Use the Law of Sines to solve the triangle. A = 110, a = 125 inches, b = 100 inches 48.23 in C  180 – 110 – 48.74 = 21.26

  7. a b A The Ambiguous Case (SSA) Given: two sides and an angle opposite to one of the given sides Three possible situations : 1) no such triangle exists; 2) one such triangle exits; 3) two distinct triangles may satisfy the conditions.

  8. A is obtuse a > b One Solution a < b No Solution b = 7 A = 114° a = 19 The Ambiguous Case (SSA)

  9. a= 9 b = 7 A = 74° The Ambiguous Case (SSA) Angle A is acute and a ≥ b One Solution

  10. A is acute a < b a > b • sin A Two Solutions a < b • sin A No Solutions a = b • sin A One Solution b a A The Ambiguous Case (SSA)

  11. C b = 20 in a = 18 in 76 A B Example 3 (SSA): Use the Law of Sines to solve the triangle. A = 76, a = 18 inches, b = 20 inches There is no angle whose sine is 1.078. There is no triangle satisfying the given conditions.

  12. C 49.8 b = 12.8 cm a =11.4 cm 72.2 58 A B c Example 4 (SSA): Use the Law of Sines to solve the triangle. A = 58, a = 11.4 cm, b = 12.8 cm 10.3 cm C  180 – 58 – 72.2= 49.8 Two different triangles can be formed. Example continues.

  13. C C1 =49.8 b = 12.8 cm a =11.4 cm 72.2 58 B1 A c1 =10.3 cm C C2 =14.2 b = 12.8 cm a = 11.4 cm 107.8 58 A B2 Example (SSA) continued: Use the Law of Sines to solve the second triangle. A = 58, a = 11.4 cm, b = 12.8 cm B2  180–72.2 = 107.8  C2 180 – 58 – 107.8= 14.2 B1 c2 = 3.3 cm

  14. Summary: Number 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.) • If sin B > 1, then no triangle satisfies the given conditions. If sin B = 1, then one trianglesatisfies the given conditions and B = 90°. If 0 < sin B < 1, then either one or two triangles satisfy the given conditions • If sin B = value, then let B1 = sin-1 value, and use B1 for B in the first triangle. • Let B2 = 180° – B1. If A + B2 < 180°, then a second triangle exists, and use B2 for B in the second triangle. If A + B2 > 180°, then there is only one triangle.

  15. C Example 5: a b Find the area of the triangle. A = 74, b = 103 inches, c = 58 inches 74 A B c Area of an Oblique Triangle 103 in 58 in

  16. Example 6 Finding the Area of a Triangular Lot Find the area of a triangular lot containing sides that measure 24 yards and 18 yards and form an angle of 80° A = ½(18)(24)sin80° A = 212.7 square yards

  17. 20 70 34 14 Example 7 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? A B Flagpole height: b 16 m C The flagpole is approximately 9.5 meters tall.

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