Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 months ago

Pre-Calculus Help?

A box has dimensions 2"x 3" x 4". (See illustration.) Determine the angle θ formed by the diagonal of the 2"x 3" side and the diagonal of the 3" x 4" side. Round your answer to the nearest degree.

Answer choices:

A. 60° 

B. 50°

C. 62°

D. 65°

Attachment image

3 Answers

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  • 8 months ago

    Another way to get this is using a vector dot product.  Let A be the vector on the 4x3 diagonal and B the the vector on the 2x3 diagonal, both originating at the corner where the vertex of θ is.  Then:

        A • B = ||A|| ||B|| cos θ

    Using Cartesian coordinates for the vectors:

        A = (4, 3, 0)

        B = (0, 3, 2)

    ...in a coordinate system with the origin at the vertex of θ, the x-axis parallel to the length 4 edges, the y-axis parallel to the length 3 edges and z parallel to the length 2 edges.  The dot product statement above then becomes:

    (4, 3, 0) • (0, 3, 2) = √(4² + 3²) √(3² + 2²) cos θ

    (4)(0) + (3)(3) + (0)(2) = √25 √13 cos θ

    9 = 5 √13 cos θ

    θ = cos⁻¹ 9 / (5 √13) πθ

    My calculator says that's about 60.051 degrees.

  • Mangal
    Lv 4
    8 months ago

    see image ....................

    Attachment image
  • If you draw a third side, you have a triangle and you can use the law of cosines.

    a = sqrt(2^2 + 3^2) = sqrt(13)

    b = sqrt(3^2 + 4^2) = sqrt(25) = 5

    c = sqrt(2^2 + 4^2) = sqrt(20) = 2 * sqrt(5)

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

    20 = 13 + 25 - 2 * 5 * sqrt(13) * cos(T)

    20 = 38 - 10 * sqrt(13) * cos(T)

    10 * sqrt(13) * cos(T) = 18

    5 * sqrt(13) * cos(T) = 9

    cos(T) = 9 / (5 * sqrt(13))

    cos(T) = 9 * sqrt(13) / 65

    T = cos-1(9 * sqrt(13) / 65)

    T = ‭60.050918050079687159403079779982‬

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