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Chapter 9: Triangle Trigonometry. L9.2 The Area of a Triangle Heron’s Formula (SSS) SAS formula (based on K=½bh) Area of a Segment of a Circle. Oblique (General) Triangles. Triangle Congruence:
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Chapter 9: Triangle Trigonometry L9.2 The Area of a Triangle Heron’s Formula (SSS) SAS formula (based on K=½bh) Area of a Segment of a Circle
Oblique (General) Triangles • 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 * Note that SSA has ambiguity and SSS can generate an illegal triangle.
Heron’s Formula, ΔABC: where s = semiperimiter Heron’s Formula: SSS • Heron’s formula determines area based on the lengths of the 3 sides of a triangle. Examples: • Find area of ΔABC if a = 9, b = 7 and c = 6. • Find area of ΔDEF if d = 3, e = 4 and f = 9. ← You do this one! This is valid triangle (7 + 6) > 9. s = (9 + 7 + 6)/2 = 11 units2 Illegal triangle!
Finding the Area of a Triangle: SAS • A triangle is uniquely determined given SAS • Given ∆ABC, let h be the lengthof its altitude from B • From geometry: K = ½bh • By right angle trig, we know that sin C = h/a or h = a·sinC • So K = ½ab·sinC • If 2 other sides and their included angle are selected, we could repeat this procedure to get 3 different formulas: K = ½ · (side1) · (side2) · (sine of included angle) Area of a Triangle, using SAS: K = ½ab·sinC = ½ac·sinB = ½bc·sinA
Area of Triangle (SAS): Examples • Find the area of this triangle: • ∆ABC has area 5cm2. Sides a, b are of length 4cm and 5cm, respectively. Find all possible measures of angle C. • Find the area of the parallelogram below. Need SAS, so included angle is 180 – (20 + 10) = 150°. K = ½·4·10·sin(150°) = 10 units2. 5 = ½·4·5·sin(C) sin(C) = 10/20 = ½ → Q1 or Q2 angle → 2 ∆s are possible C = sin-1(1/2) = 30° or 150° Creating two congruent Δs with a diagonal, K = 2KΔ = 2·½(8)(12.5)∙sin(40°) ≈ 64.3cm2
r θ r − = Segments of Circles • A segment of a circle is a region bounded by an arc of the circle and a chord connecting the endpoints of the arc. • The area of a segment can be found by subtracting area of the triangle from the area of the sector (L7.2). Ksegment = Ksector − K∆ Ksector = ½r2θ (θ in radians) K∆ = ½r2sinθ (θ in radians) Recall (7.2): Area of a Segment (θ in radians): K = ½r2θ − ½r2sinθ = ½r2(θ − sinθ)
x = 1 − = X2 + y2 = 9 r θ r Segments of Circles: Examples • Find the area of the segment of a circle with radius 7 and central angle 1.5 radians. • Graph the region satisfying both inequalities and find its area: x2 + y2 ≤ 9; x ≥ 1 K = ½r2θ − ½r2sinθ = ½r2(θ − sinθ); Need r & θ – have them! =½(7)2(1.5 – sin(1.5)) ≈ 12.3 units2 K = ½r2θ − ½r2sinθ = ½r2(θ − sinθ) – Need r & θ r = 3, θ = ? Let Φ = θ/2 cos(Φ)= ⅓, so Φ = cos-1(⅓) ~ 1.23 So, θ = 2Φ ~ 2∙1.23 ~ 2.46 Now, K ~ ½(3)2(2.46 - sin(2.46)) ≈ 8.23 units2 Φ
