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The Law

The Law. of SINES. B. a. c. A. C. b. The Law of SINES. For any triangle (right, acute or obtuse), you may use the following formula to solve for missing sides or angles:. C. a. b h. A. B. c. Proof of the Law of SINES. Prove that sin A = a sin B b

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The Law

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  1. The Law of SINES

  2. B a c A C b The Law of SINES For any triangle (right, acute or obtuse), you may use the following formula to solve for missing sides or angles:

  3. C a b h A B c Proof of the Law of SINES Prove that sin A = a sin B b In triangle CDA, sin A = h / b In triangle CDB, sin B = h/a So, sin A = h/b sin B h/a Since h/b = a / b h/a Therefore sin A = a sin B b D

  4. Use Law of SINES when ... • AAS - 2 angles and 1 adjacent side • ASA - 2 angles and their included side • SSA(this is an ambiguous case) you have 3 dimensions of a triangle and you need to find the other 3 dimensions . Use the Law of Sines if you are given:

  5. B 80° a = 12 c 70° A C b Example 1 You are given a triangle, ABC, with angle A = 70°, angle B = 80° and side a = 12 cm. Find the measures of angle C and sides b and c.

  6. B 80° a = 12 c 70° A C b Example 1 (con’t) The angles in a ∆ total 180°, so angle C = 30°. Set up the Law of Sines to find side b:

  7. B 80° a = 12 c 70° 30° A C b = 12.6 Example 1 (con’t) Set up the Law of Sines to find side c:

  8. B 80° a = 12 c = 6.4 70° 30° A C b = 12.6 Example 1 (solution) Angle C = 30° Side b = 12.6 cm Side c = 6.4 cm Note: We used the given values of A and a in both calculations. Your answer is more accurate if you do not used rounded values in calculations.

  9. B 30° c a = 30 115° C A b Example 2 You are given a triangle, ABC, with angle C = 115°, angle B = 30° and side a = 30 cm. Find the measures of angle A and sides b and c.

  10. B 30° c a = 30 115° C A b Example 2 (con’t) To solve for the missing sides or angles, we must have an angle and opposite side to set up the first equation. We MUST find angle A first because the only side given is side a. The angles in a ∆ total 180°, so angle A = 35°.

  11. B 30° c a = 30 115° C A 35° b Example 2 (con’t) Set up the Law of Sines to find side b:

  12. B 30° c a = 30 115° C A 35° b = 26.2 Example 2 (con’t) Set up the Law of Sines to find side c:

  13. B 30° c = 47.4 a = 30 115° C A 35° b = 26.2 Example 2 (solution) Angle A = 35° Side b = 26.2 cm Side c = 47.4 cm Note: Use the Law of Sines whenever you are given 2 angles and one side!

  14. The Ambiguous Case (SSA) When given SSA (two sides and an angle that is NOT the included angle) , the situation is ambiguous. The dimensions may not form a triangle, or there may be 1 or 2 triangles with the given dimensions. We first go through a series of tests to determine how many (if any) solutions exist.

  15. C = ? ‘a’ - we don’t know what angle C is so we can’t draw side ‘a’ in the right position b A B ? c = ? The Ambiguous Case (SSA) In the following examples, the given angle will always be angle A and the given sides will be sides a and b. If you are given a different set of variables, feel free to change them to simulate the steps provided here.

  16. C = ? C = ? a a b b A B ? A B ? c = ? c = ? The Ambiguous Case (SSA) Situation I: Angle A is obtuse If angle A is obtuse there are TWO possibilities If a ≤ b, then a is too short to reach side c - a triangle with these dimensions is impossible. If a > b, then there is ONE triangle with these dimensions.

  17. C a = 22 15 = b 120° A B c The Ambiguous Case (SSA) Situation I: Angle A is obtuse - EXAMPLE Given a triangle with angle A = 120°, side a = 22 cm and side b = 15 cm, find the other dimensions. Since a > b, these dimensions are possible. To find the missing dimensions, use the Law of Sines:

  18. C a = 22 15 = b 36.2° 120° A B c The Ambiguous Case (SSA) Situation I: Angle A is obtuse - EXAMPLE Angle C = 180° - 120° - 36.2° = 23.8° Use Law of Sines to find side c: Solution: angle B = 36.2°, angle C = 23.8°, side c = 10.3 cm

  19. C = ? b a A B ? c = ? The Ambiguous Case (SSA) Situation II: Angle A is acute If angle A is acute there are SEVERAL possibilities. Side ‘a’ may or may not be long enough to reach side ‘c’. We calculate the height of the altitude from angle C to side c to compare it with side a.

  20. h C = ? b a A B ? c = ? The Ambiguous Case (SSA) Situation II: Angle A is acute First, use SOH-CAH-TOA to findh: Then, compare ‘h’ to sides a and b . . .

  21. C = ? a b h A B ? c = ? The Ambiguous Case (SSA) Situation II: Angle A is acute If a < h, then NO triangle exists with these dimensions.

  22. C C b b a h a h A B A c c B The Ambiguous Case (SSA) Situation II: Angle A is acute If h < a < b, then TWO triangles exist with these dimensions. If we open side ‘a’ to the outside of h, angle B is acute. If we open side ‘a’ to the inside of h, angle B is obtuse.

  23. C b a h A B c The Ambiguous Case (SSA) Situation II: Angle A is acute If h < b < a, then ONE triangle exists with these dimensions. Since side a is greater than side b, side a cannot open to the inside of h, it can only open to the outside, so there is only 1 triangle possible!

  24. C b a = h A c B The Ambiguous Case (SSA) Situation II: Angle A is acute If h = a, then ONE triangle exists with these dimensions. If a = h, then angle B must be a right angle and there is only one possible triangle with these dimensions.

  25. C = ? a = 12 15 = b h 40° A B ? c = ? The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 Given a triangle with angle A = 40°, side a = 12 cm and sideb = 15 cm, find the other dimensions. Find the height: Since a > h, but a< b, there are 2 solutions and we must find BOTH.

  26. C a = 12 15 = b h 40° A B c The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 FIRST SOLUTION: Angle B is acute - this is the solution you get when you use the Law of Sines!

  27. C 1st ‘a’ 15 = b a = 12 40° A c B 1st ‘B’ The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse - use the first solution to find this solution. In the second set of possible dimensions, angle B is obtuse, because side ‘a’ is the same in both solutions, the acute solution for angle B & the obtuse solution for angle B are supplementary. Angle B = 180 - 53.5° = 126.5°

  28. Angle B = 126.5° Angle C = 180°- 40°- 126.5° = 13.5° C 15 = b a = 12 126.5° 40° A B c The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 1 SECOND SOLUTION: Angle B is obtuse

  29. C 13.5° C 15 = b a = 12 86.5° 15 = b a = 12 126.5° 40° A B c = 4.4 40° 53.5° A B c = 18.6 The Ambiguous Case (SSA) Situation II: Angle A is acute - EX. 1 (Summary) Angle B = 126.5° Angle C = 13.5° Side c = 4.4 Angle B = 53.5° Angle C = 86.5° Side c = 18.6

  30. C = ? a = 12 10 = b h 40° A B ? c = ? The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 2 Given a triangle with angle A = 40°, side a = 12 cm and side b = 10 cm, find the other dimensions. Since a > b, and h is less than a, we know this triangle has just ONE possible solution - side ‘a’opens to the outside of h.

  31. C a = 12 10 = b 40° A B c The Ambiguous Case (SSA) Situation II: Angle A is acute - EXAMPLE 2 Using the Law of Sines will give us the ONE possible solution:

  32. if a < b no solution if angle A is obtuse if a > b one solution (Ex I) if a < h no solution (Ex II-1) if h < a < b 2 solutions one with angle B acute, one with angle B obtuse if angle A is acute find the height, h = b*sinA (Ex II-2) if a > b > h 1 solution If a = h 1 solution angle B is right The Ambiguous Case - Summary

  33. AAS • ASA • SSA (the ambiguous case) Use the Law of Sines to find the missing dimensions of a triangle when given any combination of these dimensions. The Law of Sines

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