Chapter 9: Triangle Trigonometry

1. (SSS) Chapter 9: Triangle Trigonometry L9.4 Law of Cosines (SAS) a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C

2. Oblique Triangles (General, not just Right ∆’s) • Triangle Congruence: • A triangle can be determined uniquely if certain facts are known.Need 3 facts, one of which is a side. • SAS: Side – Angle – Side (2 sides and their included angle) • ASA: Angle – Side – Angle (2 (actually 3) angles and a side) • SSA: Side – Side – Angle (2 sides and a non-included angle)* • SSS: 3 sides* • Note: AAA is not a condition of congruence, but of similarity. • Chapter Overview: • L9.1: Solving Right Triangles • L9.3: Solving ∆s w/ Law of Sines (ASA & SSA*) – *Ambiguous Case • L9.4: Solving ∆s w/ Law of Cosines (SAS, SSS*) – *Illegal Case poss • L9.2: Area of ∆ (SAS & SSS); Area of Circle Segments • L9.5: Navigation & Surveying Applications

3. D Law of Cosines (SSS) Law of Cosines: Proof • Altitude AD of length h separates ΔABC into two right triangles with a common side AD. • Use the Pythagorean Theorem and the cosine ratios to derive a relationship among a, b, c and the measure of C. c2 = (a – x)2 + h2 Apply Pythagorean Theorem to ΔABD = a2 – 2ax + x2 + h2 = a2 – 2ax + b2 For ΔADC, b2 = x2 + h2 = a2 – 2a(bcosC) + b2 = a2 + b2 – 2ab cos C Next 2 steps replace x and h w/ a, b, c, C: Law of Cosines (SAS) a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C opp2 = adj12 + adj22 – 2(adj1)(adj2)cos(angle)

4. (SSS) Law of Cosines (SAS)* a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C • The Pythagorean Thm is a special case of the Law of Cosines: cos(90°) = 0 c2 = a2 + b2 – 2ab cos C → c2 = a2 + b2 – 2ab ·0 → c2 = a2 + b2 • 3 facts uniquely determine a triangle. • Law of Cosines requires • 2 sides and its included angle: SAS • 3 sides: SSS • Acute & obtuse angles have cosines of different signs, so no ambiguity!! • However, the SSS case can be illegal. • The triangle inequality states that for any Δ, the length of any side must be less than or equal to the sum of the other two sides. • Ex: a = 4, b = 10, c = 3 is not a Δ,because a + c < b • To check, pick the longest side and make sure the sum of the other 2 is greater. * opp2 = adj12 + adj22 – 2(adj1)(adj2)cos(angle)

5. SAS Examples Draw a sketch. • Solve the triangle a = 2, b = 3, C = 60°. • Solve the triangle B = 130°, a = 3, c = 7 ← You do this one! Find the remaining side, c, using the Law of Cosines. c2 = a2 + b2 -2ab cosC c2 = 22 + 32 – 2(2)(3)cos(60°) A = 40.9°, B = 79.1° c = 2.64 c2 = 4 + 9 – 12·½ = 7 You can use Law of Sines to find remaining angles, but care must be takendue to sine ambiguity*. Always find angle opposite smallest remaining side.Why is this? Because that angle cannot be obtuse (its not the largest one). Find angle A: A = 14.4°, C = 35.6° b = 9.22 B = 180 – (60 + 40.9) = 79.1°. * Alternatively, you can use the Law of Cosines – more computation but no ambiguity!

6. SSS Examples A = 36.3°, B = 26.4° C = 117.3° Check that this is a legal triangle: longest side, 6 < sum of other sides (4+3)? Yes! • Solve the triangle a = 4, b = 3, c = 6. • Solve the triangle w/ sides 6, 12, & 15← You do this one! Draw a sketch. Always find biggest angle first. It’s the only one that can be obtuse. Then you can use the simpler Law of Sines to find the remaining angles. Use Law of Sines to find remaining angles: Note, A + B + C = 180°, so you can find the 3rd angle, A = 180° – (117.3 + 26.4)°. Angles from largest to smallest: 108.2°, 49.5°, 22.3°

7. Solving Oblique Triangles: Summary