Clayton needs to reflect the triangle below across the line y = x. On a coordinate plane, triangle A B C has points (5.5, 7), (6
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A cube, is made off 6 squarial faces, so all faces on that cube, are squares, the front, back, left, right, top and bottom.
a square has all equal sides, and also all right angles, so all angles in a square are 90°. Let's say the sides are "x" long.
now, if we run a plane on that cube diagonally, check the picture below, the diagonal side at the bottom, by usin the 45-45-90 rule as you see it there, will be x√2.
let's keep in mind that, "x" is opposite side of that angle θ, and then x√2 will be the adjacent side of it.
and we can use those two to get the tangent and then the inverse tangent to get the value, as you see it in the picture.
if you need the angle in radians, run the inverse tangent again, just make sure your calculator is in radians mode.
a square has all equal sides, and also all right angles, so all angles in a square are 90°. Let's say the sides are "x" long.
now, if we run a plane on that cube diagonally, check the picture below, the diagonal side at the bottom, by usin the 45-45-90 rule as you see it there, will be x√2.
let's keep in mind that, "x" is opposite side of that angle θ, and then x√2 will be the adjacent side of it.
and we can use those two to get the tangent and then the inverse tangent to get the value, as you see it in the picture.
if you need the angle in radians, run the inverse tangent again, just make sure your calculator is in radians mode.