A rectangular box has dimensions 3m, 4m, 12m.
Find the cosine of the angle between the longest diagonal of the box and the face whose dimensions are 3m x 12m.
Thanks in advance
We basically need to draw a decent diagram. The "longest diagonal" is the diagonal joining two opposite corners of the box. Let the "height" be 4, width 3, and length 12, and draw it so the 3 x 12 face is on the bottom. Take say the diagonal from the "top back left" vertex of the box to the "bottom front right" vertex. Then the angle this makes to the bottom face (which is the 3 x 12 one) is found using right-angled trigonometry. The right-angled triangle we want is the one with vertices ABC, where A is the "bottom front right" vertex, B is the "top back left" vertex, and C is the "bottom back left" vertex (so BC is one of the edges of the box, of length 4 (height of box)).
Also, AC = sqrt(3^2 + 12^2), using Pythagoras. And AB (which is the "longest diagonal" of the box) has length sqrt(3^2 + 4^2 + 12^2). Calling the angle we want alpha, we have alpha = angle BAC, so cos(alpha) = AC/AB, and we know the lengths of these.
