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In rectangle \(ABCD\) below, we have \(DP = PC\) and \(\angle BAC = 28^\circ\) . If \(Q\) is the point on \(\overline{AC}\) such that \(\overline{DQ}\perp \overline{AC}\), what is \(\angle QDP\) in degrees?

 Oct 4, 2019
 #1
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 BAC  = 28°  = DCP     [ a consequence of parallels AB and CD  cut by transversal AC ]

 

And since DP = PC then angle DCP  = angle PDC

 

Therefore angle DPC  =  180 - 2(28)  =  180 - 56  = 124

 

And by the exterior angle theorem

 

Angle QPD + Angle QDP  = Angle DPC

 

    90  + Angle QDP  =  124

 

Angle QDP =  124 - 90 =  34°

 

 

 

cool cool cool

 Oct 4, 2019

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