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# hard geometry

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Find AC.

Jun 26, 2020

#1
+25644
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Find AC.

$$\text{Let AB=c }\\ \text{Let BC=a }\\ \text{Let AC=x }$$

$$\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle ABE:} \\ \hline \sin(90^\circ-\varphi) &=& \dfrac{c}{9} \\\\ \cos(\varphi) &=& \dfrac{c}{9} \\\\ \mathbf{c} &=&\mathbf{ 9\cos(\varphi)} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle ABC:} \\ \hline \sin(90^\circ-2\varphi) &=& \dfrac{c}{x} \\\\ \cos(2\varphi) &=& \dfrac{c}{x} \\\\ \mathbf{c} &=&\mathbf{ x\cos(2\varphi)} \qquad (2) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (1)=(2): & c = \mathbf{ 9\cos(\varphi)} &=& \mathbf{ x\cos(2\varphi)} \\\\ & 9\cos(\varphi) &=& x\cos(2\varphi) \\\\ & \mathbf{x} &=& \mathbf{\dfrac{9\cos(\varphi) } {\cos(2\varphi)}} \\\\ & x &=& \dfrac{9\cos(\varphi) } {\cos^2(\varphi)-\sin^2(\varphi)} \\\\ & x\Big(\cos^2(\varphi)-\sin^2(\varphi)\Big) &=& 9\cos(\varphi) \quad | \quad : \cos^2(\varphi) \\\\ & x\Big(1-\tan^2(\varphi)\Big) &=& \dfrac{9}{\cos(\varphi)} \quad | \quad \dfrac{1}{\cos(\varphi)} = \sqrt{1+\tan^2(\varphi) } \\\\ & x\Big(1-\tan^2(\varphi)\Big) &=& 9\sqrt{1+\tan^2(\varphi) } \\\\ & \mathbf{x} &=& \mathbf{\dfrac{9\sqrt{1+\tan^2(\varphi) }} {1-\tan^2(\varphi)}} \qquad (3) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle ACD:} \\ \hline \sin(45^\circ+\varphi) &=& \dfrac{a}{8\sqrt{2}} \\\\ \dfrac{\sqrt{2}}{2}\Big(\sin(\varphi)+\cos(\varphi) \Big) &=& \dfrac{a}{8\sqrt{2}} \\\\ 8\Big(\sin(\varphi)+\cos(\varphi) \Big) &=& a \\\\ \mathbf{a} &=&\mathbf{8\Big(\sin(\varphi)+\cos(\varphi) \Big)} \qquad (4) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle ABC:} \\ \hline \sin(2\varphi) &=& \dfrac{a}{x} \quad | \quad \sin(2\varphi)=\dfrac{2\tan(\varphi)}{1+\tan^2(\varphi)} \\\\ \dfrac{2\tan(\varphi)}{1+\tan^2(\varphi)} &=& \dfrac{a}{x} \\\\ \mathbf{a} &=& \mathbf{\dfrac{2x\tan(\varphi)}{1+\tan^2(\varphi)}} \qquad (5) \\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline (4)=(5): & a = \mathbf{8\Big(\sin(\varphi)+\cos(\varphi) \Big)} &=& \mathbf{\dfrac{2x\tan(\varphi)}{1+\tan^2(\varphi)}} \\\\ & 8\Big(\sin(\varphi)+\cos(\varphi) \Big) &=& \dfrac{2x\tan(\varphi)}{1+\tan^2(\varphi)} \\\\ & 4\Big(\sin(\varphi)+\cos(\varphi) \Big) &=& \dfrac{x\tan(\varphi)}{1+\tan^2(\varphi)} \quad | \quad \cos(\varphi) = \dfrac{1}{ \sqrt{1+\tan^2(\varphi)} } \\\\ & 4\left(\dfrac{1}{ \sqrt{1+\tan^2(\varphi)} } +\sin(\varphi) \right) &=& \dfrac{x\tan(\varphi)}{1+\tan^2(\varphi)} \quad | \quad \sin(\varphi) = -\dfrac{\tan(\varphi)}{ \sqrt{1+\tan^2(\varphi)} } \\\\ & 4\left(\dfrac{1}{ \sqrt{1+\tan^2(\varphi)} } -\dfrac{\tan(\varphi)}{ \sqrt{1+\tan^2(\varphi)} } \right) &=& \dfrac{x\tan(\varphi)}{1+\tan^2(\varphi)} \\\\ & \dfrac{ 4\Big( 1-\tan(\varphi) \Big) } { \sqrt{1+\tan^2(\varphi)} } &=& \dfrac{x\tan(\varphi)}{1+\tan^2(\varphi)} \\\\ & 4\Big( 1-\tan(\varphi) \Big) &=& \dfrac{x\tan(\varphi)\sqrt{1+\tan^2(\varphi)}}{1+\tan^2(\varphi)} \\\\ & 4\Big( 1-\tan(\varphi) \Big) &=& \dfrac{x\tan(\varphi)}{\sqrt{1+\tan^2(\varphi)}} \\\\ & \mathbf{x} &=& \mathbf{ \dfrac{4\sqrt{1+\tan^2(\varphi)}\Big( 1-\tan(\varphi) \Big) } {\tan(\varphi)} } \qquad (6)\\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (3)=(6): & x = \mathbf{\dfrac{9\sqrt{1+\tan^2(\varphi) }} { 1-\tan^2(\varphi)} } &=& \mathbf{ \dfrac{4\sqrt{1+\tan^2(\varphi)}\Big( 1-\tan(\varphi) \Big) } {\tan(\varphi)} } \\\\ & \dfrac{9} { 1-\tan^2(\varphi)} &=& \dfrac{4\Big( 1-\tan(\varphi) \Big) } {\tan(\varphi)} \\\\ & \dfrac{9}{4}*\tan(\varphi) &=& \Big( 1-\tan^2(\varphi)\Big) \Big( 1-\tan(\varphi) \Big) \\\\ & \dfrac{9}{4}*\tan(\varphi) &=& 1-\tan(\varphi) - \tan^2(\varphi)+\tan^3(\varphi) \\\\ & \ldots \\ & \mathbf{\tan^3(\varphi) -\tan^2(\varphi)+\dfrac{5}{4}*\tan(\varphi)+1} &=& \mathbf{0} \\ & \Rightarrow \varphi &=& \arctan(\dfrac{1}{2}) \\ &\mathbf{\varphi} &=& \mathbf{26.5650511771^\circ} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{9\cos(\varphi) } {\cos(2\varphi)}} \\\\ x &=& \dfrac{9\cos(26.5650511771^\circ) } {\cos(2*26.5650511771^\circ)} \\\\ x &=& \dfrac{8.04984471900} {0.6} \\\\ \mathbf{x} &=& \mathbf{13.4164078650} \\ \hline \end{array}$$

Jun 26, 2020
edited by heureka  Jun 26, 2020