Question

A rectangular box has length 2 inches, width 8 inches, and a height of 10 inches. Find the angle between the diagonal of the box and the diagonal of its base. The angle should be measured in radians.

Answer 1

The angle between the diagonal of the box and the diagonal of its base is approximately 0.153 radians.

Step-by-step explanation:

To find the angle between the diagonal of the box and the diagonal of its base, we can use trigonometry.

The diagonal of the box can be found using the Pythagorean theorem:

D = sqrt(l^2 + w^2 + h^2)

D = sqrt(2^2 + 8^2 + 10^2)

D = sqrt(4 + 64 + 100)

D = sqrt(168)

D ≈ 12.9618 inches

The diagonal of the base can be found using the Pythagorean theorem as well:

D_base = sqrt(l^2 + w^2)

D_base = sqrt(2^2 + 8^2)

D_base = sqrt(4 + 64)

D_base = sqrt(68)

D_base ≈ 8.2462 inches

Now we can find the angle between the two diagonals using the cosine formula:

cos(angle) = (D_base^2 + D^2 - d^2) / (2 * D_base * D)

cos(angle) = (8.2462^2 + 12.9618^2 - 10^2) / (2 * 8.2462 * 12.9618)

cos(angle) = (67.9847 + 168.5222 - 100) / (2 * 67.9847)

cos(angle) = 136.5069 / 135.9694

angle ≈ 0.153 radians

Answer 2

a) diagonal box = 12.9 in

b) diagonal base = 8.2 in

Step-by-step explanation:

w = 8 in

h = 10 in

L = 2 in

required:

a) diagonal of the box

b) diagonal of its base

referring into the attached image

a) the diagonal of the box = sqrt ( w² + h² + L²)

diagonal box = sqrt (8² + 10² + 2²)

diagonal box = 12.9 in

b) diagonal of its base = sqrt ( w² + L²)

diagonal base = sqrt ( 8² + 2²)

diagonal base = 8.2 in