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In the diagram below, we have \(\overline{AB}\parallel\overline{CD}\),\(QP = QR\) ,\(\angle BPQ = x^\circ + 100^\circ\) , and \(\angle APR = x^\circ\). Find \(x\).

https://latex.artofproblemsolving.com/2/3/6/23637a6b4f6218e50583ee5b39bcc517771d75a8.png
 

 Feb 5, 2023
 #1
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x = 144/5

 Feb 5, 2023
 #2
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Angle APR==Angle PRQ ==x


Angle PRQ ==Angle QPR==x  {Because triangle PQR is an isosceles triangle}


Angle BPQ ==Angle PQR ==100 + x 


From isosceles triangle PQR, we have:


(100 + x) + x + x ==180


3x ==180 - 100 ==80 


x ==80 / 3 ==26 + 2/3 == 26.6667 degrees

 Feb 5, 2023

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