Question

A rectangular box has length 20 inches, width 20 inches, and a height of 2 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

Answer:

0.0707 radians

Step-by-step explanation:

Given that:

The dimension of the rectangular box is:

length (l) = 20 inches

width (w) = 20 inches

height (h) = 2 inches

The length of the diagonal of the rectangular box is :

$D_1 = \sqrt{ l^2+w^2+h^2}$

$D_1 = \sqrt {(20)^2+(20)^2+(2)^2}$

$D_1 = \sqrt {400+400+4}$

$D_1 = \sqrt {804}$

$D_1 = 28.35$

The length of the diagonal of the base of the rectangular box is:

$D_2 = \sqrt{l^2+w^2}$

$D_2 = \sqrt{(20)^2+(20)^2}$

$D_2 = \sqrt{400+400}$

$D_2 = \sqrt{800}$

$D_2 = 28.28$

If we take a critical look  at a rectangular box; we will realize that the diagonal of the base is the adjacent side & diagonal of the box is the hypotenuse of a triangle formed by the rectangular box.

Therefore, the angle between them is :

$Cos \theta = \frac{Diagonal \ of \ the \ base \ }{Diagonal \ of \ the \ box }$

$Cos\ \theta = \frac{D_2}{D_1 }$

$Cos\ \theta = \frac{28.28}{28.35 }$

$Cos\ \theta = 0.9975$

$\theta = Cos^{-1} \ 0.9975$

$\theta = 4.052^0$  to radians

$\theta = 0.0707 \ radians$