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.