6.20 The Cosine Law.notebook - nowyoudothemath.weebly.com...6.20 The Cosine Law.notebook 3 May 29,...

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6.20 The Cosine Law.notebook 1 May 29, 2020 Solutions b 2 =a 2 +c 2 2accos(B) b 2 = 12 2 + 10 2 2(12)(10)cos(72) b 2 = 169.835... b = √(169.835...) b = 13.03... b = 13 cm c 2 =a 2 +b 2 2abcos(C) c 2 = 20 2 + 17 2 2(20)(17)cos(39) c 2 = 160.540... c = √(160.540...) c = 12.670... c = 13 mm

Transcript of 6.20 The Cosine Law.notebook - nowyoudothemath.weebly.com...6.20 The Cosine Law.notebook 3 May 29,...

  • 6.20 The Cosine Law.notebook

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    May 29, 2020

    Solutions

    b2 = a2 + c2  2accos(B)

    b2 = 122 + 102  2(12)(10)cos(72)

    b2 = 169.835...

    b = √(169.835...)

    b = 13.03...

    b = 13 cm

    c2 = a2 + b2  2abcos(C)

    c2 = 202 + 172  2(20)(17)cos(39)

    c2 = 160.540...

    c = √(160.540...)

    c = 12.670...

    c = 13 mm

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    d2 = e2 + f2  2efcos(D)

    d2 = 3.22 + 4.42  2(3.2)(4.4)cos(66)

    d2 = 18.146...

    d = √(18.146...)

    d = 4.2598...

    d = 4.3 m

    z2 = x2 + y2  2xycos(Z)

    z2 = 1.82 + 2.22  2(1.8)(2.2)cos(48)

    z2 = 2.7804...

    z = √(2.7804...)

    z = 1.6675...

    z = 1.7 cm

    p2 = m2 + n2  2mncos(P)

    p2 = 6.22 + 4.82  2(6.2)(4.8)cos(54)

    p2 = 26.4950...

    p = √(26.4950...)

    p = 5.1473...

    p = 5.1 mm

    u2 = t2 + v2  2tvcos(U)

    u2 = 1.82 + 1.42  2(1.8)(1.4)cos(52)

    u2 = 2.0970...

    u = √(2.0970...)

    u = 1.4481...

    u = 1.4 cm

    d2 = e2 + f2  2efcos(D)

    d2 = 1.12 + 1.62  2(1.1)(1.6)cos(74)

    d2 = 2.7997...

    d = √(2.7997...)

    d = 1.6732...

    d = 1.7 km

    52Ο 74Ο

    1.8 cm

    1.4 cm 1.1 km

    1.6 km

    D

    E

    FTU

    V

    u d

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    r2 = p2 + q2  2pqcos(R)

    r2 = 132 + 152  2(13)(15)cos(70)

    r2 = 260.6121...

    r = √(260.6121...)

    r = 16.143

    r = 16 cm

    sin(P)    p

    = sin(R)r

     sin(P)   13

    = sin(70)16

    sin(P) =  13sin(70)16sin(P) = 0.7635...

    P = sin1(0.7635...)P = 49.7737...

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    a2 = b2 + c2  2bccos(A)

    a2 = 82 + 92  2(8)(9)cos(71)

    a2 = 98.118...

    a = √(98.118...)

    a = 9.9055...

    a = 10 m

    sin(B)    b

    = sin(A)a sin(B)    8

    = sin(71)10

    sin(B) =  8sin(71)10sin(B) = 0.7564...

    B = sin1(0.7564...)

    B = 49.149...

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    x2 = w2 + y2  2wycos(X)

    x2 = 102 + 112  2(10)(11)cos(80)

    x2 = 182.7974...

    x = √(182.7974...)

    x = 13.52...

    x = 14 cm

    sin(W)    w

    = sin(X)x sin(W)    10

    = sin(80)14

    sin(W) =  10sin(80)14sin(W) = 0.7034...

    W = sin1(0.7034...)

    W = 44.703...

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    b2 = h2 + s2  2hscos(B)

    b2 = 2.02 + 2.52  2(2.0)(2.5)cos(70)

    b2 = 6.8297...

    b = √(6.8297...)

    b = 2.6134...

    b = 2.6 km

    sin(H)    h

    = sin(B)b sin(H)   2.0

    = sin(70)2.6

    sin(H) =  2.0sin(70)2.6sin(H) = 0.7228...H = sin1(0.7228...)H = 46.289...

    a)

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    a2 = b2 + c2  2bccos(A)

    a2 = 9152 + 15252  2(915)(1525)cos(37)

    a2 = 934059.950...

    a = √(934059.950...)

    a = 966.47...A

    B

    C

    37Ο1525 m

    915 m

    a) b)

    The runways are 966 metres apart

    a

    a2 = b2 + c2  2bccos(A)

    a2 = 102 + 102  2(10)(10)cos(104.5)

    a2 = 250.076...

    a = √(250.076...)

    a = 15.814...

    a = 15.8 cm

    104.5Ο 10 cm10 cm

    A

    BC

    a)a

    b) Both bond lengths are 10 cm which makes the triangle isosceles.