# Geometry/Trigonometry Question on two congruent isosceles triangles inside and outside a square. [closed]

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.

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

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