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I redid the diagram and added to it. $C$ is the center of the circle that forms the arch, $r$ is the radius of the circle, $E$ is the midpoint of the line segment $AB,$ and $y = H  h.$ I assume your measurements are in inches. The first task is to find the value of $r.$ Triangle $AEC$ is a right triangle so Pythagoras Theorem gives \[CA^2 = AE^2 + EC^2 \mbox{ or } r^2 = \left( \frac{w}{2} \right)^2 + (ry)^2\] solving this expression for $r$ and simplifying gives \[r = \frac{4 y^2 + w^2}{8 y} \mbox{ units.}\]
I next need to know the measure of the angle $ACB.$ Again from triangle $AEC,$ \[\sin \left( ACE \right) = \frac{AE}{CA} = \frac{w/2}{r}\] and hence the measure of angle $ACE$ is \[\sin^{1}\left( \frac{w/2}{r} \right).\] $sin^{1}$ is sometimes written $\arcsin$ or ASIN. Thus the measure of the angle $ACB,$ I'm going to call it $\theta,$ is \[\theta = 2 \times \sin^{1}\left( \frac{w/2}{r} \right).\]
The length $a$ of an arc of a circle of radius $r$ and central angle $\theta$ radians is given by \[a = r \theta.\]
The area of the share is the area of the rectangular part, $w \times h,$ plus the area of the circular cap $ADBC.$ The area of the circular cap is the area of the sector $ADBC$ minus the area of the triangle $ABC.$ The area of the sector is \[\frac{1}{2} r^2 \theta\] and the area of the circular cap $ADCB$ is \[ \frac{1}{2} r^2 \theta  \frac{1}{2} w \times (r  y)\] Hence the area of the shape is \[w \times h + \left( \frac{1}{2} r^2 \theta  \frac{1}{2} w \times (r  y) \right)\]
I used an Excel Spread Sheet to verify my arithmetic I I have attached it for your use if you desire. For a different example just change the values of $H, h,$ and $w.$ Harley Richard wrote back again
Glad to help when I can. I modified the diagram again by inserting a coordinate system with $C$ the origin and the Yaxis through $E.$ The line $PQ$ is tangent to the circle at $B.$ I think the angle you want is the angle $QBF.$ Triangle $EBC$ is a right angled triangle, the length of $EB$ is w/2 and the length of $BC$ is $r$ and hence by Pythagoras Theorem \[CE^2 = BC^2  EB^2 \mbox{ or } CE = \sqrt{r^2  \left(\frac{w}{2}\right)^2}.\] Hence the coordinates of $B$ are $\left( w/2, \sqrt{r^2  {w^2}/4}\right).$ From this the slope of the line $CB$ is \[\frac{\sqrt{r^2  {w^2}/4}}{w/2}\] The key to this question is that the lines $QP$ and $CB$ are perpendicular since $CB$ is a radius of the circle an $QP$ is tangent to the circle at $B.$ Since $QP$ is perpendicular to $CB$ its slope is the negative reciprocal of the slope of $CB$ and hence \[\mbox{ slope of QP is } \frac{w/2}{\sqrt{(r^2  {w^2}/4)}}.\] But the slope of $QP$ is the tangent of the angle $QBE$ and hence \[\mbox{angle QBE} = \tan^{1}\left(\frac{w/2}{\sqrt{(r^2  {w^2}/4})}\right)\] where this time I am going to calculate the measure of the agile in degrees. Finally the measure of the angle $QBF$ is $90^{o} + \mbox{ angle } QBE.$ I have modified the spreadsheet to include the angle measurement and attached it for your use. If you want $H$ to be 69 inches just change the $H$ value in the spreadsheet to 69. Let me know if this is not the angle you want or if I can be of any further assistance. Harley




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