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# Similar Triangles

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The question is right here in the picture below Oct 7, 2018
edited by shreyas1  Oct 7, 2018

#1
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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 \(\angle {PQC}\).

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

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

\(\boxed {m\angle {PQC} = 15^\circ}\) .
Oct 8, 2018
#2
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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   Oct 8, 2018
edited by CPhill  Oct 8, 2018
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Thank you so much guys.

Oct 8, 2018