Is Trigonometry Mandatory?

Note that by double angle identities ,

$\tan 30^\circ = \dfrac1{\sqrt3} = \dfrac{2\tan15^\circ}{1-\tan^2 15^\circ } \Rightarrow \tan 15^\circ = 2-\sqrt3 \; .$

The figure above shows that

$FG = FP + PG = a$

$PF = \dfrac a2 \cdot \tan 15^\circ = \dfrac a2 (2 -\sqrt3)$

$PG = a - PF = a - a \left( 1 - \dfrac{\sqrt 3}2 \right) = a \dfrac{\sqrt 3} 2$

$PC^2 = \left( \dfrac a2\right)^2 + \left( \dfrac {a\sqrt 3} 2 \right)^2 = a^2$

Since the two triangles $PAD$ and $BPC$ are congruent (SAS), then $PD = PC = a$ .

So the triangle $PCD$ is an equilateral triangle, thus $\angle PCD = 60^\circ$ .

Without loss of generality, assume the square is a unit square. As the angles of $\Delta PAB$ are given and one of its side lengths is known, the other side lengths can be found. Now, use sine rule to $\Delta PBC$ and find angle $PCB$ which turns out to be $30^\circ$ . Hence, the answer is $60^\circ$ .

Let angle PCD =x AB/sin150 =PB/sin15 and BC/sin(15+x) =PB/sin(90-x) sin150/sin15 =sin(15+x)/cos x We get tan x =3^(1/2) x=60