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Trigonometry Part II Math 416

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Game Plan Area of Triangles Area of Triangles Traditional Traditional Sine Sine Heros Heros Sine Law Sine Law Two examples Two examples Word Problem Word Problem

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

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

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

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

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Using Heros to find Area of Triangle Now a totally different approach was found by Hero or Heron Now a totally different approach was found by Hero or Heron His approach is based on perimeter of a triangle His approach is based on perimeter of a triangle

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

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

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

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

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

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

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

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

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2 nd Example Eg Eg y 75° θ x 18 Complete the Triangle Lets get θ 1 st x = 18 Sin 75 Sin 62 x = 19.69… now y? y = 18 Sin 43 Sin 62 y = 13.90 43° θ = 180 – 75 = 62… x?

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

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Division ÷ 1 1 ÷ 1 = 1 2 ÷ 1 = 2 3 ÷ 1 = 3 4 ÷ 1 = 4 5 ÷ 1 = 5 6 ÷ 1 = 6 7 ÷ 1 = 7 8 ÷ 1 = 8 9 ÷ 1 = 9 10 ÷ 1 = 10 11 ÷ 1 = 11 12 ÷ 1 = 12 ÷ 2 2 ÷ 2 =

Division ÷ 1 1 ÷ 1 = 1 2 ÷ 1 = 2 3 ÷ 1 = 3 4 ÷ 1 = 4 5 ÷ 1 = 5 6 ÷ 1 = 6 7 ÷ 1 = 7 8 ÷ 1 = 8 9 ÷ 1 = 9 10 ÷ 1 = 10 11 ÷ 1 = 11 12 ÷ 1 = 12 ÷ 2 2 ÷ 2 =

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