Find the length PD in the figure below. For reference:
Find the length PD in the figure below.
ABCDEF is a regular hexagon
AB = 10cm
AQE is a sector( center at F)
PD is tangent of sector

My progresss:
$\triangle DEH:10.sin60^o = \frac{DF}{2} \therefore 
DF = 10\sqrt3$
$\triangle DFQ :DF^2 = DQ^2+10^2 \implies DQ = \sqrt{300-100} \therefore DQ = 10\sqrt2  $
I can't find PQ

 A: Not very "elegant", but with trigonometry you get
$$\measuredangle PDB = \frac{\pi}3-\arcsin \frac1{\sqrt 3}$$
Thus
\begin{eqnarray}
\overline{PD} &=&\frac{10\sqrt 3}{\cos\left(\frac{\pi}3-\arcsin \frac1{\sqrt 3}\right)}=\\
&=&\frac{10\sqrt 3}{\frac12 \cdot \sqrt{\frac23}+\frac{\sqrt{3}}2\cdot \frac1{\sqrt 3}}=\\
&=&\frac{10 \sqrt 3}{\frac1{\sqrt 6}+\frac12}=60\cdot(\sqrt 3-\sqrt 2).
\end{eqnarray}
A: We can, without loss of generality, set up a coordinate system so that the center of the hexagon is at (0, 0) and the top and bottom are parallel to the x-axis.  Then A= (-10 sin(30), -10 cos(30)), B= (10 sin(30), -10 cos(30)), C= (10, 0), D= (10 sin(30), 10 cos(30)), E= (-10 sin(30), 10 cos(30)), and F= (-10, 0).
The circle with center at F and radius 10 is given by $(x+ 10)^2+ y^2= 100$ and any line through D is $y= m(x- 10 sin(30))+ 10 cos(30)= m(x- 5)+ 5\sqrt{3}$.  For the circle, $2(x+ 10)+ 2yy'= 0$ so $y'= -\frac{x+ 10}{y}$.  A line tangent to the circle at (x, y) must have slope $m= -\frac{x+ 10}{y}$.
We have $y= \frac{x+ 10}{y}(x- 5)+ 5\sqrt{3}= \frac{x^2+ 5x- 50}{y}+ 5\sqrt{3}$.  The point at which the line DP is tangent to the circle must satisfy
$y= \frac{x+ 10}{y}(x- 5)+ 5\sqrt{3}= \frac{x^2+ 5x- 50}{y}+ 5\sqrt{3}$ and $(x+ 10)^2+ y^2= 100$. Solve those two equations for x and y, the coordinates of the line of tangency.
Once you have that, you can find the equation of the line through point D and this point of tangency, then find the coordinates of P, where the line intersects $y= -10 cos(30)= -5\sqrt{30}$ and so find the distance between D and P.
