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Given an ellipse with dimensions 165 x 135, how can I find the angle α as shown in the image? All distances from the blue circular arc to point M are 50 units.

enter image description here

Using the equation $F=\sqrt{(y/2)^2-(x/2)^2}$ I found the distance between foci points and the centre of the ellipse to be 47.43, however I am not sure how to appropiately use this value.

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Let $H$ be the projection of $P$ on the major axis and $O$ the ellipse center. Then: $$ PH=D\sin\alpha,\quad OH=a-D\cos\alpha, $$ where $a=165/2$ m is the semi major axis. Plug these into the ellipse equation: $OH^2/a^2+PH^2/b^2=1$ (where $b=135/2$ m is the semi minor axis) to get an equation for $\cos\alpha$: $$ {D^2(a^2-b^2)\over a^2b^2}\cos^2\alpha+2{D\over a}\cos\alpha-{D^2\over b^2}=0. $$

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  • $\begingroup$ How can the final equation be rearranged so that cos α is the subject and thus inverse cos can then be used to determine angle α? $\endgroup$ – Nikolai Sas Aug 30 '16 at 15:46
  • $\begingroup$ Do you know how to solve a 2nd degree equation? If so, set $\cos\alpha=x$, find the solution $x$ of the equation (you must pick the right one!) and finally $\alpha=\arccos x$. $\endgroup$ – Aretino Aug 30 '16 at 17:01
  • $\begingroup$ As it is not a right angled triangle, I do not understand how I can use the right angle trig functions of PH = D sin α and OH = a - D cos α? Is the fact it is an isosceles triangle relevant? $\endgroup$ – Nikolai Sas Sep 9 '16 at 2:17
  • $\begingroup$ Probably my answer wasn't detailed enough: $H$ is the projection of $P$ on the vertical axis, the one with $M$ as an endpoint, and $PMH$ is thus a right triangle. $\endgroup$ – Aretino Sep 9 '16 at 13:36

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