In triangle XYZ, we have \(\angle X = 60^\circ\) and \(\angle Y = 45^\circ\). The bisector of \(\angle X\) intersects \(\overline{YZ} \) at point D, and DX = 24. What is the area of triangle XYZ?
Using the angle bisector theorem, the area of triangle XYZ works out to 108 + 144*sqrt(3).
+26367 In triangle \(XYZ\), we have \(\angle X = 60^\circ\) and \(\angle Y = 45^\circ\).
The bisector of \(\angle X\) intersects \(\overline{YZ}\) at point D, and \(DX = 24\).
What is the area of triangle \(XYZ\)?
sin-rule:
\(\begin{array}{|rcll|} \hline \dfrac{\sin(30^\circ)}{u} &=& \dfrac{\sin(45^\circ)}{24} \\\\ \mathbf{u} &=& \mathbf{\dfrac{24 \sin(30^\circ)}{\sin(45^\circ)}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{\sin(30^\circ)}{v} &=& \dfrac{\sin(75^\circ)}{24} \\\\ \mathbf{v} &=& \mathbf{\dfrac{24 \sin(30^\circ)}{\sin(75^\circ)}} \\ \hline \end{array}\)
\(\mathbf{u+v = \ ?}\)
\(\begin{array}{|rcll|} \hline u+v &=& \dfrac{24 \sin(30^\circ)}{\sin(45^\circ)} + \dfrac{24 \sin(30^\circ)}{\sin(75^\circ)} \\\\ u+v &=& 24 \sin(30^\circ) \left( \dfrac{1}{\sin(45^\circ)} + \dfrac{1}{\sin(75^\circ)} \right) \\ && \boxed{\sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\ \sin(75^\circ) = \dfrac{1+\sqrt{3}} {2\sqrt{2}} \\ \sin(30^\circ) = \dfrac{1}{2} } \\ u+v &=& \dfrac{24}{2} \left( \dfrac{2} {\sqrt{2}} + \dfrac{2\sqrt{2}}{1+\sqrt{3}} \right) \\ u+v &=& 24 \left( \dfrac{1} {\sqrt{2}} + \dfrac{ \sqrt{2}}{1+\sqrt{3}} \right) \\ \ldots \\ \mathbf{u+v} &=& \mathbf{12\sqrt{6}} \\ \mathbf{\left(u+v\right)^2} &=& \mathbf{864} \\ \hline \end{array}\)
sin-rule:
\(\begin{array}{|rcll|} \hline \dfrac{\sin(60^\circ)}{u+v} &=& \dfrac{\sin(75^\circ)}{z} \\\\ \mathbf{z} &=& \mathbf{\dfrac{(u+v) \sin(75^\circ)}{\sin(60^\circ)}} \\ \hline \end{array}\)
The area of triangle \(XYZ\):
\(\begin{array}{|rcll|} \hline A &=& \dfrac{z(u+v)\sin(45^\circ)}{2} \quad | \quad \mathbf{z=\dfrac{(u+v) \sin(75^\circ)}{\sin(60^\circ)}} \\\\ A &=& \dfrac{ \left(u+v\right)^2\sin(45^\circ)\sin(75^\circ)}{2\sin(60^\circ)} \quad | \quad \mathbf{\left(u+v\right)^2=864} \\\\ A &=& \dfrac{864\sin(45^\circ)\sin(75^\circ)}{2\sin(60^\circ)} \\\\ A &=& \dfrac{432\sin(45^\circ)\sin(75^\circ)}{\sin(60^\circ)} \\\\ && \boxed{\sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\ \sin(75^\circ) = \dfrac{1+\sqrt{3}} {2\sqrt{2}} \\ \sin(60^\circ) = \dfrac{\sqrt{3}}{2} } \\ A &=& \dfrac{432\sqrt{2}}{2} \dfrac{(1+\sqrt{3})} {2\sqrt{2}} \dfrac{2}{\sqrt{3}} \\\\ A &=& 216 \dfrac{(1+\sqrt{3})} {\sqrt{3}} \\\\ A &=& 216 \dfrac{(1+\sqrt{3})} {\sqrt{3}} \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ A &=& \dfrac{216}{3} (1+\sqrt{3})\sqrt{3} \\\\ A &=& 72(1+\sqrt{3})\sqrt{3} \\\\ A &=& 72(3+\sqrt{3}) \\ \mathbf{A} &=& \mathbf{340.707658145 } \\ \hline \end{array}\)
