A geometry problem by Fahim Shahriar Shakkhor

Draw perpendiculars $PM$ and $PN$ on $AB$ and $CD$ respectively.

Let the side of the square be $x$ . Then, $PM+PN=x$

In $\triangle PBM$ ,

$\tan PBM = \frac{PM}{BM}$

$\Rightarrow \tan 15^{\circ} = \frac{x-PN}{\frac{x}{2}}$

$\Rightarrow 2 - \sqrt{3} = \frac{2(x-PN)}{x}$

$\Rightarrow PN = \frac{\sqrt{3}x}{2}$

.

In $\triangle PCN$ ,

$\Rightarrow \tan PCN = \frac{PN}{CN}$

$\Rightarrow \tan PCN = \frac{\frac{\sqrt{3}x}{2}}{\frac{x}{2}}$

$\Rightarrow \tan PCN = \sqrt{3}$

$\angle PCN = 60^{\circ}$

$\triangle APD$ and $\triangle BPC$ are congruent. So $\angle PCD = \angle PDC = \angle CPD = \boxed {60^{\circ}}$

This solution is best explained with a diagram (the $\LaTeX$ which produced it can be found here )

We begin by adding interior points $Q,R,S$ in addition to $P$ in a symmetrical manner, so that: $\angle QBC=\angle QCB=\angle RCD=\angle RDC=\angle SDA=\angle SAD=15^\circ,$ and by symmetry one can see that $PQRS$ is a square and triangles $ASP,BPQ,CQR,DRS$ are equilateral.

With this symmetrical construction, by examining the angles around the point S, one can see that APD is an isosceles triangle with paired sides the same size as the length of the square. One thus sees by symmetry $PD=PC=DC$ and thus that the triangle $PDC$ is an isosceles triangle. Thus $\angle CPD = 60^\circ$ .

Let side of square be $a$ .

Construct a line through P parallel to AD meeting AB at Q and CD at R. $\implies QR = a$

$\triangle PQA \cong \triangle PQB$ ... by SAA test $\implies QA = QB = a/2$ $\implies PQ = \frac{a}{2} \times tan(15^0)$

Let $\angle CPD = \theta$ . So in $\triangle CPR, \angle CPR = \frac{\theta}{2}$

$\implies tan(\frac{\theta}{2}) = \frac{CR}{PR} = \frac{CR}{QR-PQ} = \frac{\frac{a}{2}}{a - \frac{a}{2} \times tan(15^0)} = \frac{1}{2 - tan(15^0)} = \frac{1}{2 - (2 - \sqrt{3}) } = \frac{1}{\sqrt{3}}$

$\implies \frac{\theta}{2} = 30^0 \implies \theta = 60^0$

Answer: $\boxed{60}$

Let length of square is x

Triangle PAB is an isosceles triangle.Draw a perpendicular bisector PZ from P on AB .

tan 15= PZ/(x/2)=2PZ/x

PZ= xtan15/2

Notice triangle PDC is also an isosceles triangle.

The perpendicular bisector from P to DC = x-(xtan15/2)

tan (PDC)= (2x-xtan15)/x

2-tan15=tan (PDC)

tan (PDC)=\sqrt{3}

angle PDC=60=angle PCD

So angle CPD= 180-120=60

I constructed the whole stuff and measured angle(CPD)....because i was not able to solve..

assume that each side of the square is 10,
draw a perpendicular from P onto AB at E,
Angle APB = 180 - (15+15) = 150
Hence angle APE = 150/2 = 75
Now, Applying Sine Rule, in triangle APE ,we get PE = 1.34
Draw a line Parallel to AB and CD passing through P, meeting AD at F and BC at G
Hence FA = GB = PE = 1.34 ; DF = CG = 10 - 1.34 = 8.66 ; FP = PG = 5
Triangles DFP and CGP are right angled triangles, So DP = CP = 10
Applying Sine Rule again, we get , angle FPD = angle GPC = 60
Therefore, angle CPD = 180 - (60 + 60) = 60