What's the measure of the angle formed by a diagonal of the box with the base of the box?

Question

A rectangular box has dimensions of length=6, width=4, and height=5. What's the measure of the angle formed by a diagonal of the box with the base of the box?

Answer 1

first draw a diagonal on the bottom rectangle and find its length

call it d

d^2=6^2+4^2

d^2=36+16

d^2=52

d=√52

d=7.2111

the diagonal of the box, the diagonal of the bottom base, and the height form a right triangle with the diagonal of the box being the hypotenuse

call the diagonal of the box b

b^2=5^2+7.2111^2

b^2=25+52

b^2=77

b=√77

b=8.774964

use the sin or cos to find the angle

sinθ=5/8.774964

sinθ=0.5698029

using arcsin on your calculator or a table or whatever...

sin^-1θ=0.5698029

θ=34º 44' 11" or 34.73648º (approximately)

using cos...

cosθ=7.2111/8.774964

cosθ=0.8217811

using arcos..

cos^-1θ=0.8217811

θ=34º 44' 11" or 34.73651º (approximately)