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Default rotating an Isoceles Triangle in the shape of a Gauge needle

Hi,

does anyone know how to rotate an Isosceles triange (in the shape
of a Speedometer needle) from 0 to 100 (the speedometer dial is 240
deg in shape).

i am basically finding the x & y to center the Isosceles needle in
the center of the speedometer by taking the LEFT, WIDTH & TOP of the
speedometer image.
x=shp.Left+Shp.Width/2
y=shp.Top+Shp.Height/2

so the above would remain fixed for centering the Isosceles
triangle.

but how to calculate the Angle or the new X, Y positions for the
point traversing the Speedometer Arc which is 240 degrees in shape
based on a cell values ranging from 0 to 100?

any help would be most appreciated.

Please see attached file.