# 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°

### 3 Answers

- husoskiLv 71 month 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.

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- 1 month ago
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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- Captain Matticus, LandPiratesIncLv 71 month agoReport
I figured my choice of numbers would show you. Do you see the back wall that measures 2" x 4"? That's where. It connects the other ends and forms a triangle.

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