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 The question is right here in the picture below

 

shreyas1  Oct 7, 2018
edited by shreyas1  Oct 7, 2018
 #1
avatar+28 
+1

Okay. Let's work this through step by step. If \(m\angle {BAC} = 70^\circ\), and \(\triangle{ABC} \sim \triangle {PAQ}\), then \(\angle{APQ}\) should also be \(70^\circ\).

 

If \(\angle{APQ} = 70^\circ , \angle{CPQ} = 110^\circ\).

 

And since \(\triangle{QPC} \sim \triangle{ABQ} , \angle{AQB} = 110^\circ\).

 

Also, since \(\angle{AQB} = 110^\circ , \angle{AQC} = 70^\circ\).

 

Okay, now for the second part.

 

Since \(\triangle {ABQ} \sim \triangle {QPC} , \angle {PQC} = \angle {BAQ}\).

 

Let both these angles be \(x\).

 

This means that \(\angle {QAP} = \angle {AQP}\), since \(\angle {BAC} = 70^\circ\) and \(\angle{AQC} = 70^\circ\), and both \(\angle {QAP}\)  and \(\angle{AQP}\) equal \((70-x)^\circ\).

 

Since \(\angle {APQ} = 70^\circ\), both  \(\angle {QAP}\)  and \(\angle{AQP}\) are equal to \(55^\circ\).

 

Now the final step. Now that we have the angle measurements of \(\angle {AQB}\) and \(\angle {AQP}\), we can solve for \(\angle {PQC}\).

\(\angle {AQB} + \angle {AQP} = 165^\circ\).

 \(\angle {PQC} = 180^\circ - 165^\circ = 15^\circ\)

\(\boxed {m\angle {PQC} = 15^\circ}\)

laugh

KnockOut  Oct 8, 2018
 #2
avatar+89874 
+1

Since triangle ABC  is similar to triangle PAQ

Then angle BAC  = angle APQ

But angle BAC  = 70

So angle APQ  = 70

And angle QPC  is supplemental to angle APQ

So angle QPC  = 110

 

And triangle ABQ  is simialr to triangle QCP

So  angle ABQ = angle QCP  

But angle ABQ  = angle ABC = angle QCP = angle ACB

 

So...triangle ABC  isosceles  with angles ABC, ACB  being equal

So...their measures  =  [ 180  - m angle BAC ] / 2  =   [180  - 70] / 2  = 110 / 2  = 55

And since m angle ACB  = m angle QCP...then m angle QCP  = 55

 

So...in triangle QPC......m angle PQC  = [ 180  - m angle QPC - m angle QCP ] =

 

180  - 110  - 55  = 180 - 165  =   15°  =  m angle PQC

 

 

cool cool cool

CPhill  Oct 8, 2018
edited by CPhill  Oct 8, 2018
 #3
avatar+460 
0

Thank you so much guys.

shreyas1  Oct 8, 2018

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