+288 A circle has center \(O\) and radius \(6.\) Points \(A, B, C,\) and \(D\) are chosen on the circle such that \(AD=BC,\) and such that rays \(\overrightarrow{AD}\) and \(\overrightarrow{BC}\) meet at point \(P\) outside the circle. Suppose that \(AB=10\) and \(\angle DOC=90\)°. Compute \(AP.\)
-----------Thanks! 
+26367 A circle has center \(O\) and radius \(6\).
Points \(A,\ B,\ C,\ \) and \(D\) are chosen on the circle such that \(AD=BC\),
and such that rays \(\overrightarrow{AD}\) and \(\overrightarrow{BC}\) meet at point \(P\) outside the circle.
Suppose that \(AB=10\) and \(\angle DOC=90^\circ\). Compute \(AP\).
\(\begin{array}{|rcll|} \hline \text{In $\triangle AEO$ } \\ \hline \mathbf{\sin(A)} &=& \mathbf{\dfrac{5}{6}} \\ \mathbf{A} &=& \mathbf{56.4426902381^\circ} \\ \hline \end{array} \begin{array}{|rcll|} \hline A+B+45^\circ &=& 180^\circ \\ \mathbf{B} &=& \mathbf{135^\circ-A} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline \text{In $\triangle DFO$ } \\ \hline FD&=& 6\sin(45^\circ) \quad | \quad \sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\ FD&=& \dfrac{6\sqrt{2}}{2} \\ \mathbf{FD} &=& \mathbf{3\sqrt{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline 2C+2\left(\dfrac{B}{2}+45^\circ \right) &=& 180^\circ \\ 2C+B+90^\circ &=& 180^\circ \\ 2C+B&=& 90^\circ \\ 2C &=& 90^\circ-B \quad | \quad B=135^\circ-A \\ 2C &=& 90^\circ-(135^\circ-A) \\ 2C &=& A-45^\circ \\ \mathbf{C} &=& \mathbf{\dfrac{A-45^\circ}{2}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \text{In $\triangle DFP$ } \\ \hline \sin(C) &=& \dfrac{FD}{x} \quad | \quad \mathbf{FD=3\sqrt{2}} \\\\ \sin(C) &=& \dfrac{3\sqrt{2}}{x} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin(C)} \quad | \quad \mathbf{C=\dfrac{A-45^\circ}{2}} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin\left(\dfrac{A-45^\circ}{2}\right)} \quad | \quad A=56.4426902381^\circ \\\\ x &=& \dfrac{3\sqrt{2}}{\sin\left(\dfrac{ 56.4426902381^\circ-45^\circ}{2}\right)} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin\left(\dfrac{ 11.4426902381^\circ}{2}\right)} \\\\ x &=& \dfrac{3\sqrt{2}}{\sin(5.72134511904^\circ)} \\\\ x &=& \dfrac{3\sqrt{2}}{0.09969044344} \\\\ x &=& \dfrac{4.24264068712}{0.09969044344} \\\\ \mathbf{x} &=& \mathbf{42.5581484121} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In $\triangle ADO$ } \\ \hline \mathbf{AD} &=& \mathbf{6^2+6^2-2*6*6*\cos(B)} \\\\ AD &=& 72\Big(1 - \cos(B)\Big) \quad | \quad \mathbf{B=135^\circ-A} \\ AD &=& 72\Big(1 - \cos(135^\circ-A)\Big) \\ AD &=& 72\Big(1 - \cos(180^\circ-45^\circ-A)\Big) \\ AD &=& 72\Big(1 - \cos(180^\circ-(45^\circ+A)\Big) \\ AD &=& 72\Big(1 + \cos( 45^\circ+A)\Big) \\ AD &=& 72\Big(1 + \cos(101.442690238^\circ)\Big) \\ \mathbf{AD} &=& \mathbf{57.7160876877} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{AP} &=& \mathbf{x + AD} \\\\ AP &=& 42.5581484121 + 57.7160876877 \\ \mathbf{AP} &=& \mathbf{100.274236100} \\ \hline \end{array}\)
\(\mathbf{AP\approx 100.3} \)
