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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?

 

 Feb 10, 2020
 #1
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-1

Using the angle bisector theorem, the area of triangle XYZ works out to 108 + 144*sqrt(3).

 Feb 10, 2020
 #3
avatar+531 
+2

I'm getting different result; triangle area  A ≈ 340.71 u²  indecision

Triangles  XMD and XND are identical.

 

Dragan  Feb 10, 2020
edited by Dragan  Feb 10, 2020
 #2
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+1

Thanks!

 Feb 10, 2020
 #4
avatar+24389 
+4

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 \(XYZ\):

\(\begin{array}{|rcll|} \hline A &=& \dfrac{z(u+v)\sin(45^\circ)}{2} \quad | \quad \mathbf{z=\dfrac{(u+v) \sin(75^\circ)}{\sin(60^\circ)}} \\\\ A &=& \dfrac{ \left(u+v\right)^2\sin(45^\circ)\sin(75^\circ)}{2\sin(60^\circ)} \quad | \quad \mathbf{\left(u+v\right)^2=864} \\\\ A &=& \dfrac{864\sin(45^\circ)\sin(75^\circ)}{2\sin(60^\circ)} \\\\ A &=& \dfrac{432\sin(45^\circ)\sin(75^\circ)}{\sin(60^\circ)} \\\\ && \boxed{\sin(45^\circ) = \dfrac{\sqrt{2}}{2} \\ \sin(75^\circ) = \dfrac{1+\sqrt{3}} {2\sqrt{2}} \\ \sin(60^\circ) = \dfrac{\sqrt{3}}{2} } \\ A &=& \dfrac{432\sqrt{2}}{2} \dfrac{(1+\sqrt{3})} {2\sqrt{2}} \dfrac{2}{\sqrt{3}} \\\\ A &=& 216 \dfrac{(1+\sqrt{3})} {\sqrt{3}} \\\\ A &=& 216 \dfrac{(1+\sqrt{3})} {\sqrt{3}} \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ A &=& \dfrac{216}{3} (1+\sqrt{3})\sqrt{3} \\\\ A &=& 72(1+\sqrt{3})\sqrt{3} \\\\ A &=& 72(3+\sqrt{3}) \\ \mathbf{A} &=& \mathbf{340.707658145 } \\ \hline \end{array}\)

 

laugh

 Feb 10, 2020
 #5
avatar+109561 
+2

Very nice, heureka  !!!

 

 

cool cool cool

CPhill  Feb 10, 2020
 #6
avatar+531 
+1

Impressive!!!  smiley

Dragan  Feb 10, 2020
 #7
avatar+24389 
+1

Thank you, CPhill !

 

laugh

heureka  Feb 11, 2020
 #8
avatar+24389 
+1

Thank you, Dragan !

 

laugh

heureka  Feb 11, 2020

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