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# transversal and isoceles triangle problem

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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.

Mar 15, 2020

#1
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Since QP = QR, 2x + 100 = 180.  Then x = 40.

Mar 15, 2020
#2
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BPQ + QPR + APR = 180

so  x+100+QPR+x = 180

or  2x + QPR = 80 ...(1)

QRP = APR

QPR = QRP (because triangle QRP is isosceles with QR = QP) hence QRP = APR = x

use this in (1) to get:

2x + x  = 80

3x = 80

x = 80/3

Mar 15, 2020