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# In triangle XYZ, we have and . The bisector of intersects at point D, and DX = 24. What is the area of triangle XYZ?

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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
-1

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

Feb 10, 2020
#3
+531
+2

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

Triangles  XMD and XND are identical.

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

Thanks!

Feb 10, 2020
#4
+24389
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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 $$XYZ$$?

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 $$XYZ$$:

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

Feb 10, 2020
#5
+109561
+2

Very nice, heureka  !!!

CPhill  Feb 10, 2020
#6
+531
+1

Impressive!!!

Dragan  Feb 10, 2020
#7
+24389
+1

Thank you, CPhill !

heureka  Feb 11, 2020
#8
+24389
+1

Thank you, Dragan !

heureka  Feb 11, 2020