It should be fairly trivial to change the rotation point to another corner, in my current case I just need it to pivot around the upper-left.

B4X:

```
' Rectangle should be the unrotated drawbitmap target rect for the image. Left/top are where to draw it and right/bottom should be the left/top plus the image original width/height.
' Degrees is the amount to rotate it by. The returned RECT is where you need to draw using the Canvas.DrawBitmapRotated function.
Private Sub RotateRectUL(Rectangle As Rect, Degrees As Float) As Rect
Dim pfX As Float
Dim pfCX As Float
Dim pfY As Float
Dim pfCY As Float
Dim pfTempX As Float
Dim pfTempY As Float
Dim pfTheta As Float
Dim pfRotatedX As Float
Dim pfRotatedY As Float
Dim poResult As Rect
pfX = Rectangle.Left
pfY = Rectangle.Top
pfCX = pfX + (Rectangle.Width / 2) ' center of rect is the default rotation point
pfCY = pfY + (Rectangle.Height / 2)
' translate origin point
pfTempX = pfX - pfCX
pfTempY = pfY - pfCY
' calculate rotation
pfTheta = (Degrees * cPI / 180) ' convert to radians
pfRotatedX = pfTempX * Cos(pfTheta) - pfTempY * Sin(pfTheta)
pfRotatedY = pfTempX * Sin(pfTheta) + pfTempY * Cos(pfTheta)
' translate origin back to upper left
pfX = pfRotatedX + pfCX - Rectangle.Width + Rectangle.Height
pfY = pfRotatedY + pfCY - Rectangle.Height
' return the new rect
poResult.Initialize(pfX, pfY, pfX + Rectangle.Width, pfY + Rectangle.Height)
Return poResult
End Sub
```