# Finding the angles of the start and endpoint of an arc

I have a line $100$ $mm$ long and I want to draw an arc from endpoint to endpoint with a height of 3mm.

I use this formula to find the radius of the arc $$\frac{W . W}{ 8 H} + \frac{H} {2}= 418.1667.$$

But the CAD software I'm using, FreeCAD, uses radius and the angles for the start and endpoints to draw the arc. How do I find the start and end angles that will give me a 100mm straight line between them?

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See the picture below. From Pythagoras, $R^2=50^2+(R-3)^2,0=2500-6R+9,R=\frac {2509}6$ confirming your value. Then $\sin \theta=\frac {300}{2509}, \theta = \arcsin \frac{300}{2509}\approx 6.867^\circ$. Since the tangent is perpendicular to the radius at the point of tangency, this is the angle between the tangent and the chord, which I suspect is the angle you are looking for.

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I don't recognize which formula this one comes from 0=2500−6R+9 and I don't follow where you got 300/2509. I was trying 90 - sin(418.1667/415.1667) = ~6.243, but that wasn't coming out right. – getSurreal Jul 22 '13 at 20:56
It helps if I do my formula correct in excel and understand when to use arcsin instead of sin. asin(418.1667/415.1667)*180/PI(). Now I get 6.867. – getSurreal Jul 22 '13 at 21:05