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Trigonometry Part II

Trigonometry Part II . Math 416. Game Plan . Area of Triangles Traditional Sine Hero’s Sine Law Two examples Word Problem. Traditional Area of Triangles. We are now going to be working with other triangles besides right angle triangles The 1 st concept we will look at is area

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Trigonometry Part II

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  1. Trigonometry Part II Math 416

  2. Game Plan • Area of Triangles • Traditional • Sine • Hero’s • Sine Law • Two examples • Word Problem

  3. Traditional Area of Triangles • We are now going to be working with other triangles besides right angle triangles • The 1st concept we will look at is area • Area of triangle = half the base x height • A = ½ b h or A = bh 2

  4. Using Sin for Area of Triangle • We note the base is a side of the triangle • The height must be at 90° or perpendicular to that base (not up and down) • Eg Note b =8 h =9 A = ½ bh A = ½ (8)(9) A = 36 8 9

  5. Sin • Now let us try and get a different perspective. • Consider We are given two sides and the contained angle. From trig Sin θ = h p h = p Sin θ From formula A = ½ bh A = ½ qp Sinθ p θ Create formula for A q

  6. Another Example Find Area 12 A = ½ (18)(12) Sin 47° A = 78.99 47° 18

  7. Using Hero’s to find Area of Triangle • Now a totally different approach was found by Hero or Heron • His approach is based on perimeter of a triangle

  8. Be My Hero and Find the Area P = a + b + c (perimeter) • Consider p = a + b + c / 2 or p = P / 2 (semi perimeter) a A = p (p-a) (p-b) (p-c) b Hence, by knowing the sides of a triangle, you can find the area c

  9. Be My Hero and Find the Area P = 9 + 11 + 8 = 28 • Eg p = 14 A = p (p-a) (p-b) (p-c) 9 11 A = 14(14-9)(14-11)(14-8) A = 14 (5) (3) (6) A = 1260 8 A = 35.5

  10. Be My Hero and Find the Area P = 42 + 43 + 47 • Eg p = 66 A = p (p-a) (p-b) (p-c) 42 43 A = 66(24)(23)(19) A = 692208 47 A = 831.99

  11. Be My Hero and Find the Area P = 9 + 7 + 3 • Eg p = 9.5 A = p (p-a) (p-b) (p-c) 9 7 A = 9.5(0.5)(2.5)(6.5) A = 77.19 3 A = 8.79

  12. Sin Law With respect to Angle A • Consider Sin A = h/b h = b sin A C With respect to Angle B Sin B = h/a h = a sin B b a h Thus, aSinB = bSinA Divide both sides by ab B A Sin B = Sin A b a

  13. Sin Law • Now we can do this again using Angle C • What we get is the Sin Law for side lengths • a = b = c . Sin A Sin B Sin C Sin Law for angles • Sin A = Sin B = Sin C a b c. Notice when getting angles Sin on TOP (think a on top)

  14. Notes • Each expression is actually 3 formulae • You do not need the whole thing • Always look for the Side – Angle - Combo

  15. 1st Example • Eg Complete the Triangle Let’s get angle or θ 1st β 8 15 Sin θ = Sin 57 8 15 57° θ x θ = 27°… now Beta β= 180 – 57 – 27 = 96°… now x x = 15 Sin 96 Sin 57 x = 17.79

  16. 2nd Example • Eg Complete the Triangle Let’s get θ 1st θ x θ = 180 – 75 = 62… x? y x = 18 Sin 75 Sin 62 75° 43° 18 x = 19.69… now y? y = 18 Sin 43 Sin 62 y = 13.90

  17. Word Problem x = 200 Sin 73 Sin 63 x = 214.66 • A surveyor creates the following map • What is the shortest distance if Billy goes from home to school, to the post office and home? Billy’s House y = 200 Sin 44 Sin 63 y = 155.93 200m Dist = 214.66+155.93+200 = 570.59 m 63° 73° School Post Office

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