1. Law of Sines Trigonometry MATH 103 S. Rook

2. Overview Sections 7.1 & 7.2 in the textbook: – Law of Sines: AAS/ASA Case – Law of Sines: SSA Case 2

3. Law of Sines: AAS/ASA Case

4. Oblique Triangles Oblique Triangle: a triangle containing no right angles All of the triangles we have studied thus far have been right triangles – We can apply SOHCAHTOA or the Pythagorean Theorem only to right triangles Naturally most triangles will not be right triangles – Thus we need a method to apply to find side lengths and angles of other types of triangles 4

5. Four Cases for Oblique Triangles By definition a triangle has three sides and three angles for a total of 6 components – We can find the measure of all sides and all angles if we know AT LEAST 3 of these components Broken down into four cases: – AAS or ASA Measure of two angles and the length of any side – SSA Length of two sides and the measure of the angle opposite one of two known sides Known as the ambiguous case because none, one, or two triangles could be possible 5

6. Four Cases for Oblique Triangles (Continued) – SSS Length of all three sides – SAS Length of two sides and the measure of the angle opposite the third (possibly unknown) side – AAA is NOT a case because there are an infinite number of triangles that can be drawn Recall that the largest side is opposite the largest and angle, but there is no limitation on the length of the side! 6

7. Law of Sines The first two cases (AAS/ASA and SSA) are covered by the Law of Sines: – i.e. the ratio of the measure of any side of a triangle to its corresponding angle yields the same constant value This constant value is different for each triangle – The Law of Sines can be proved by dropping an altitude from an oblique triangle and using trigonometric functions with the right angle See page 339 ALWAYS draw the triangle! 7

8. Law of Sines: AAS/ASA (Example) Ex 1: Use the Law of Sines to solve the triangle – round answers to two decimal places: a)A = 102.4°, C = 16.7°, a = 21.6 b)A = 55°, B = 42°, c = ¾ 8

9. Law of Sines: SSA Case

10. SSA – the Ambiguous Case Occurs when we know the length of two sides and the measure of the angle opposite one of two known sides – e.g. a, b, A and b, c, C are SSA cases – e.g. a, b, C and b, c, A are NOT (they are SAS cases) To solve the SSA case: – Use the Law of Sines to calculate the missing angle across from one of the known sides – There are three possible cases: 10

11. SSA – the Ambiguous Case (Continued) The known angle corresponds to one of the given sides of the triangle (e.g. In a, b, A, the known angle is A) – i.e. the side which both the angle and length are given is the side with the known angle The missing angle corresponds to the second given side (e.g. In a, c, C, the missing angle is A) Case I: sin missing > 1 – e.g. sin missing = 1.3511 – Recall the domain for the inverse sine: -1 ≤ x ≤ 1 – NO triangle exists If -1 ≤ missing ≤ 1, at least one triangle is guaranteed to exist Inverse sine will give the value of missing in Q I 11

12. SSA – the Ambiguous Case (Continued) An angle of a triangle can have a measure of up to 180° Sine is also positive in Q II (90° < θ < 180°) meaning that it is POSSIBLE for missing to assume a value in Q II Find this second possible value using reference angles Case II: second + known ≥ 180° – i.e. the measure of the possible second angle yields a second triangle with contradictory dimensions – ONE triangle exists Case III: second + known < 180° – i.e. the measure of the possible second angle yields a second triangle with feasible dimensions – known is the same in BOTH triangles – TWO triangles exist 12

13. SSA – the Ambiguous Case (Example) Ex 2: Use the Law of Sines to solve for all solutions – round to two decimal places: a)A = 54°, a = 7, b = 10 b)A = 98°, a = 10, b = 3 c)C = 27.83°, c = 347, b = 425 d)B = 58°, b = 11.4, c = 12.8 13

14. Law of Sines Application (Example) Ex 3: A man standing near a radio station antenna observes that the angle of elevation to the top of the antenna is 64°. He then walks 100 feet further away and observes that the angle of elevation to the top of the antenna is 46° (see page 345). Find the height of the antenna to the nearest foot. 14

15. Summary After studying these slides, you should be able to: – Apply the Law of Sines in solving for the components of a triangle or in an application problem – Differentiate between the AAS/ASA and SSA cases Additional Practice – See the list of suggested problems for 7.1 & 7.2 Next lesson – Law of Cosines (Section 7.3) 15