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In triangle XYZ, XY = XZ.  W is on side XZ so that XW = WY = YZ.  Find angle XYW.

 Jun 17, 2020
 #1
avatar+307 
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**This explanation is really confusing to read without an image. Draw a picture and mark out the equal sides. Label the angles as you read the solution. 

 

Call angle XYW = a. Because XW = WY, YXZ = a.

Then, from the exterior angle theorem, ZWY = 2a. Because WY = YZ, angle WZY = 2a

Since the sum of all the angles of a triangle is \(180^\circ\), WYZ = 180-4a

Then, since XY = XZ, 180-4a+a = 2a.

Solve for this to find a = 36

 Jun 17, 2020
 #2
avatar+25565 
+1

In triangle \(XYZ\), \(XY = XZ\)\(W\) is on side \(XZ \) so that \(XW = WY = YZ\).

Find \(\angle XYW\).


\(\text{Let $\angle XYW=x$} \)

 

\(\begin{array}{|rcll|} \hline y &=& \dfrac{180^\circ-x}{2} \\ \mathbf{y} &=& \mathbf{90^\circ-\dfrac{x}{2}} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \text{In }\triangle YWZ \\ \hline 180^\circ &=& y + y + (y-x) \\ 180^\circ &=& 3y -x \quad | \quad \mathbf{y=90^\circ-\dfrac{x}{2}} \\ 180^\circ &=& 3\left(90^\circ-\dfrac{x}{2} \right)-x \\ 180^\circ &=& 270^\circ-\dfrac{3x}{2} -x \\ 180^\circ &=& 270^\circ-\dfrac{5x}{2} \\ \dfrac{5x}{2} &=& 270^\circ-180^\circ \\ \dfrac{5x}{2} &=& 90^\circ \\ x &=& \dfrac{2*90^\circ}{5} \\ x &=& 2*18^\circ \\ \mathbf{x} &=& \mathbf{36^\circ} \\ \hline \end{array}\)

 

laugh

 Jun 18, 2020

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