A box is 3 inches by 4 inches by 1 foot. Find the sine of the angle that a diagonal of the box makes with its?

d = √(3^2 + 4^2 + 1^2) = √26

cos A = 4/√26 ≈ 0.7844645...

The angle is approximately 38.3288° ≈ 38° 19' 44"

√(3² + 4²) / √((3² + 4²) + 12²)
= √(9 + 16) / √(9 + 16 + 144)
= √25 / √169
= 5/13

Okay this one's difficult to answer without a diagram.

But you should have a diagram with an oblong/vertical box. top is 4 inches by 3 inches, and the box is 1 foot long (ie. 12 inches)

The idea is basically you cut the box vertically along the diagonal downwards, then draw a diagonal on the cross section face.

So i.) the diagonal of the top end of the box is 5 inches (work out using pythagoras)

ii.) then you should have the cross section of the "cut" as a vertical rectangle which is 5 inches by 12 inches. The question is asking you to find the angle of the diagonal of this face.

Again, using Pythagoras, this diagonal is 13 inches.

Do you know SOH CAH TOA? Using this, sin theta = 12/13 (the opposite of the right angle divided by the hypotenuse.

ANS: sin theta = 12/13.

PS: Ed, it was 1 FOOT, not 1 inch.