1. H.Melikian/12001 7.1 The Sines Law Dr.Hayk Melikyan/ Departmen of Mathematics and CS/ melikyan@nccu.edu 1.Determining if the Law of Sines Can Be Used to Solve an Oblique Triangle 2.Using the Law of Sines to Solve the SAA Case or ASA Case 3.Using the Law of Sines to Solve the SSA Case 4.Using the Law of Sines to Solve Applied 5.Problems Involving Oblique Triangles

2. H.Melikian/12002 The Law of Sines If A, B, and C are the measures of the angles of any triangle and if a, b, and c are the lengths of the sides opposite the corresponding angles, then

3. H.Melikian/12003 The Law of Sines The following information is needed to use the Law of Sines. 1. The measure of an angle must be known. 2. The length of the side opposite the known angle must be known. 3. At least one more side or one more angle must be known.

4. H.Melikian/12004 Determining of the Law of Sines Can Be Used to Solve an Oblique Triangle Decide whether or not the Law of Sines can be used to solve each triangle. a.b.c. No; We are not given the measure of any angle. Yes; we are given an angle and the measure of its opposite side and an additional angle.

5. H.Melikian/12005 SolutionWe begin by drawing a picture of triangle ABC and labeling it with the given information. The figure shows the triangle that we must solve. We begin by finding B. A  B  C  180º The sum of the measurements of a triangle’s interior angles is 180º. Add. 83.5º  B  180º 50º  B  33.5º  180º A = 50º and C = 33.5º. B  96.5º Subtract 83.5º from both sides. a c AB C b  33.5º 50º Text Example (ASA)  Solve triangle ABC if A  50º, C  33.5º, and b  76.

6. H.Melikian/12006 Solve triangle ABC if A  50º, C  33.5º, and b  76. Solution Keep in mind that we must be given one of the three ratios to apply the Law of Sines. In this example, we are given that b  76 and we found that B  96.5º. Thus, we use the ratio b /sin B, or 76 /sin96.5º, to find the other two sides. Use the Law of Sines to find a and c. The solution is B  96.5º, a  59, and c  42. Find a: This is the known ratio. a c AB C b  33.5º 50º Text Example cont. Find c:

7. H.Melikian/12007 Solving a SAA Triangle Using the Law of Sines Solve the given oblique triangle. Round lengths to one decimal place. AnglesSides A = 99 a = ? B = ?? b = ? C = 50 c = 14

8. H.Melikian/12008 Solving a SAA Triangle Using the Law of Sines Solve the given oblique triangle. Round lengths to one decimal place. Find a : Find b : AnglesSides A = 99 a = ? B = 31 b = ? C = 50 c = 14

9. H.Melikian/12009 Solving a ASA Triangle Using the Law of Sines Solve oblique triangle ABC if B = 42 , A = 57 , and c = 18.6 cm. AnglesSides A = 57 a = ? B = 42 b = ? C = c = 18.6

10. H.Melikian/120010 Two Triangles No Triangle b a A h = b sin A a is less than h and not long enough to form a triangle. b A a h = b sin A a = h and is just the right length to form a right triangle. aa b A h = b sin A a is greater than h and a is less than b. Two distinct triangles are formed. b a A a is greater than h and a is greater than b. One triangle is formed. The Ambiguous Case (SSA) One Right Triangle One Triangle

11. H.Melikian/120011

12. H.Melikian/120012 Solving a SSA Triangle Using the Law of Sines ( Quiz No) Solve the triangle; if possible. AnglesSides A = ? a = 20 B = ? b = C = 50 c = 10 Since there is no angle A for which sin A > 1, there can be no triangle with the given measurements.

13. H.Melikian/120013 Solving a SSA Triangle Using the Law of Sines (One Triangle) Solve the triangle; if possible. AnglesSides A = 40 a = 30 B = ? b = 20 C = ? c = ? Two possible angles:

14. H.Melikian/120014 Solving a SSA Triangle Using the Law of Sines (One Triangle)- -cont Solve the triangle; if possible. AnglesSides A = 40 a = 30 B = 25.4 b = 20 C = 114.6 c = ?

15. H.Melikian/120015 Solving a SSA Triangle Using the Law of Sines (Two Triangles) Solve the triangle; if possible. AnglesSides A = 35 a = 60 B = ? b = 80 C = ? c = ? Two possible angles: There are two possible triangles.

16. H.Melikian/120016 Solving a SSA Triangle Using the Law of Sines (Two Triangles) Solve the triangle; if possible. AnglesSidesAnglesSides A = 35 a = 60 A = 35 a = 60 B = 49.9 b = 80 B = 130.1 b = 80 C = c = C = c =

17. H.Melikian/120017 Solving a SSA Triangle Using the Law of Sines (Two Triangles) Solve the triangle; if possible. AnglesSidesAnglesSides A = 35 a = 60 A = 35 a = 60 B = 49.9 b = 80 B = 130.1 b = 80 C = c = C = c =

18. H.Melikian/120018 Solving an Applied Problem Martin wants to measure the distance across a river. He has made a sketch. Find the distance across the river, a. a

19. H.Melikian/120019 Determining the Distance a Ship is from Port A ship set sail from port at a bearing of N 53  E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74  E. Draw a diagram. Find angle A.

20. H.Melikian/120020 Determining the Distance a Ship is from Port-cont A ship set sail from port at a bearing of N 53  E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74  E. AnglesSides A = 21 a = 69 B = b = C = c = 63 Two possible angles:

21. H.Melikian/120021 Determining the Distance a Ship is from Port-cont A ship set sail from port at a bearing of N 53  E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74  E. AnglesSides A = 21 a = 69 B = 139.9 b = C = 19.1 c = 63 The ship is 124 miles from port to point C.