Points A and B are on side \(\overline{YZ}\) of rectangle WXYZ such that \( \overline{WA}\) and \(\overline{WB}\) trisect \(\angle ZWX\). If $BY = 3$ and $AZ = 6$, then what is the area of rectangle $WXYZ$?
+26367 Points \(A\) and \(B\) are on side \(\overline{YZ}\) of rectangle \(WXYZ\) such that \(\overline{WA}\) and \(\overline{WB}\) trisect \(\angle ZWX\).
If \(BY = 3\) and \(AZ = 6\), then what is the area of rectangle \(WXYZ\)?
\(\begin{array}{|lrcll|} \hline (1) & \tan(A) &=& \dfrac{6}{y} \\\\ (2) & \tan(2A) &=& \dfrac{y}{6} \\\\ \hline (1)\times (2): & \tan(A)\times \tan(2A) &=& \dfrac{6}{y}\times \dfrac{y}{6} \\\\ & \tan(A)\times \tan(2A) &=& 1 \quad | \quad \tan(2A)=\dfrac{2\tan(A)} {1-\tan^2(A)} \\ & \tan(A)\times\dfrac{2\tan(A)} {1-\tan^2(A)} &=& 1 \\\\ & \dfrac{2\tan^2(A)} {1-\tan^2(A)} &=& 1 \\\\ & 2\tan^2(A) &=& 1-\tan^2(A) \\ & 3\tan^2(A) &=& 1 \\\\ & \tan^2(A) &=& \dfrac{1}{3} \\\\ & \tan(A) &=& \mathbf{ \dfrac{1}{\sqrt{3}} } \\ \hline (1) & \tan(A) &=& \dfrac{6}{y} \\\\ & \dfrac{1}{\sqrt{3}} &=& \dfrac{6}{y} \\\\ & \sqrt{3} &=& \dfrac{y}{6} \\\\ & \mathbf{y} &=& \mathbf{6\sqrt{3}} \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline (3) & \tan(A) &=& \mathbf{\dfrac{y}{x-3}} \\\\ & x-3 &=& \dfrac{y}{\tan(A)} \\\\ & x &=& 3+\dfrac{y}{\tan(A)} \\\\ & x &=& 3+\dfrac{6\sqrt{3}}{\dfrac{1}{\sqrt{3}}} \\\\ & x &=& 3+ 6\sqrt{3}\sqrt{3} \\\\ & x &=& 3+ 18 \\\\ & \mathbf{x} &=& \mathbf{21} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \text{The area of rectangle }~ WXYZ &=& xy \\ &=& 21*6\sqrt{3} \\ &=& \mathbf{126\sqrt{3}} \\ &=& \mathbf{218.238401754} \\ \hline \end{array}\)
