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In a square ABCD, point P is chosen inside ABCD and Q outside ABCD such that APB and BQC are congruent isosceles triangles with angle APB and BQC both equal to 80 degrees. T is a point where BC and PQ meet. What is the size of the angle BTQ?

I have been using the sine and cosine equations to try and figure it out in terms of he length of the square. I am just puzzled, how can you solve a triangle AAA (as in triangle APB has the 3 angles 50,50,80) but no sides. Thanks for any help.

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    $\begingroup$ No, sine or cosine calculations are necessary. What can you deduce about triangle PBQ. $\endgroup$
    – Doug M
    Jul 27, 2018 at 23:53
  • $\begingroup$ Triangle PBQ has an angle at B which is equal to 90 degrees so it is a right angled triangle. $\endgroup$
    – Tom Allen
    Jul 28, 2018 at 0:07
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    $\begingroup$ @Okay, but there is more. How does $PB$ relate to $BQ?$ $\endgroup$
    – Doug M
    Jul 28, 2018 at 0:15
  • $\begingroup$ PB is equal to BQ (it is an isocoles right-angled triangle) $\endgroup$
    – Tom Allen
    Jul 28, 2018 at 0:17
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    $\begingroup$ Okay.... So now you know the measure of two of the angles in triangle $QBT$ and should have no problem finding the 3rd. $\endgroup$
    – Doug M
    Jul 28, 2018 at 0:19

1 Answer 1

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Since $BQ=BP$ and $\measuredangle QBP=90^{\circ}$, we obtain $\measuredangle BPT=45^{\circ}$ and $$\measuredangle BTQ=\measuredangle BPT+\measuredangle PBT=45^{\circ}+40^{\circ}=85^{\circ}.$$

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