A farmer plans to build a triangular fence with side lengths of 500 m, 461 m, and 408 m. Determine the measures of the angles?

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  • 2 months ago

    Let the lengths of the three sides of the triangle be: 

    a = 500, b = 461 and c = 408

    The angle A (opposite to side a) may be obtained using the cosine rule as follows

    a^2 = b^2 + c^2 - 2 b c cos (A) 

    500^2 = 461^2 + 408^2 - 2 (461)(408) cos (A)

    Solve for cos (A)

    cos(A) =  -128985 / -376176 =  0.34288471353

    A = arccos(0.34288471353) = 69.9472758 degrees

    The angle B (opposite to side b) may be obtained using the cosine rule as follows

    b^2 = a^2 + c^2 - 2 a c cos (B) 

    461^2 = 500^2 + 408^2 - 2 (500)(408) cos (B) 

    cos(B) = -203943 / -408000 = 0.49986029411

    B = arccos(0.49986029411) = 60.0092424 dgrees

    angle C may be found using the fact that the sum of all angles in a triangle is 180 degrees

    A + B + C = 180

    C = 180 - 69.9472758 - 60.0092424 = 50.0434818 degrees

    more at 

    https://www.analyzemath.com/Geometry/cosine_law_pr...

  • 2 months ago

    Cosine Rule:

    c^2 = a^2 + b^2 - 2ab*cos(C)

    500^2 = 461^2 + 408^2 - 2*461*408*cos(C)

    250000 = 212521 + 166464 - 376176*cos(C)

    376176*cos(C) = 212521 + 166464 - 250000

    376176*cos(C) = 128985

    cos(C) = 128985 /376176

    cos(C) = 42995/125392

    Since C must be between 0 and pi radians (0° to 180°):

    C = arccos(42995/125392)

    C =~ 1.22 radians (69.95°)

    Use the same method for the other two angles.

  • 2 months ago

    A farmer plans to build a triangular fence 

    with side lengths of 500 m, 461 m, and 408 m. Determine the measures of the angles?

    A: 69.95°

    B: 60.01°

    C: 50.04°

  • Anonymous
    2 months ago

    I do not help people appear smarter than they actually are.

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  • 2 months ago

    Answer:

    ∠A ≈ 70°

    ∠B ≈ 60°

    ∠C ≈ 50°

    See the details in the link below.

  • 2 months ago

    Answer:

    A=70°, B=60°, C=50°

    Law of Cosines

    c² = a² + b² - 2ab cos C

    where C denotes the angle contained between sides of lengths a and b.

    Knowing all sides, the angle between a and b is

    cos C = (a² + b² - c²) / (2ab)

    C = arccos [(a² + b² - c²) / (2ab)]

    Let

    a=500

    b=461

    c=408

    and angles A, B  and C are opposite to sides a, b and c, respectively.

    C = arccos [(500² + 461² - 408²) / (2*500*461)] ≈ 50°

    Law of Sines

    c / sin C = b / sin B

    B = arcsin ((b/c) * sin C)

    B = arcsin ((461/408) * sin 50°) ≈ 60°

    Naturally, A = 180°- B - C = 180°- 60° - 50° = 70°

    Attachment image
  • Law of cosines:

    500^2 = 461^2 + 408^2 - 2 * 461 * 408 * cos(C)

    461^2 = 500^2 + 408^2 - 2 * 500 * 408 * cos(B)

    408^2 = 500^2 + 461^2 - 2 * 500 * 461 * cos(A)

    500^2 - (461^2 + 408^2) = -2 * 461 * 408 * cos(C)

    461^2 + 408^2 - 500^2 = 2 * 461 * 408 * cos(C)

    (461^2 + 408^2 - 500^2) / (2 * 461 * 408) = cos(C)

    arccos((461^2 + 408^2 - 500^2) / (2 * 461 * 408)) = C

    arccos((500^2 + 408^2 - 461^2) / (2 * 500 * 408)) = B

    arccos((500^2 + 461^2 - 408^2) / (2 * 500 * 461)) = A

    Make sure your calculator is in degree mode

    Make sure you use brackets appropriately.

    arccos(t) is also cos-1(t) on your calculator.

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