Calculating extensions in elastic string Could anyone please help me with the following question?

Here's my failed line of reasoning so far:
My diagram:

You can see that I am assuming that P moves down to lie vertically under it's original position, is even this correct?
Otherwise:
I've tried a few ways, here's the most naive, thanks for any help:
Let extension in AP = $x_1$ and extension in PB be $x_2$ then:
$L_1=\frac{3}{20\,cos\,\theta}$
and so
$x_1=\frac{3}{20\,cos\,\theta}-\frac{3}{20}$
and
$L_2=\frac{1}{20\,sin\,\theta}$
and so
$x_2=\frac{1}{20\,sin\,\theta}-\frac{1}{20}$
Hence:
$\frac{x_1}{x_2}=\frac{\frac{60-60\,cos\,\theta}{400\,cos\,\theta}}{\frac{20-20\,sin\,\theta}{400\,sin\,\theta}}=\frac{60-60\,cos\,\theta}{\cos\theta}\times\frac{sin\,\theta}{20-20\,sin\,\theta}$
which can be simplified to:
$\frac{(3-3\,cos\,\theta)\,sin\,\theta}{(1-sin\,\theta)\,cos\,\theta}$
Which isn't the required answer. Thanks for any help.
 A: We have
$$
|AP|_0 = 15\\
|BP|_0 = 5\\
r = \frac{|AP|_0+|BP|_0}{2}\\
|AP| = 2r\cos\theta\\
|BP| = 2r\sin\theta
$$
then
$$
\frac{|AP|-|AP|_0}{|BP|-|BP|_0} = \frac{20 \cos (\theta )-15}{20 \sin (\theta )-5} = \frac{4 \cos (\theta )-3}{4 \sin (\theta )-1}
$$
the solution follows as in 
Calculation of the modulus of elasticity of a stretched string
NOTE
with $\lambda$ the elastic modulus, 
$$
\lambda\left(\frac{|AP|-|AP|_0}{|AP|_0}\right)\cos\theta = \lambda\left(\frac{|BP|-|BP|_0}{|BP|_0}\right)\sin\theta
$$
(In equilibrium the horizontal projection for $F_{AP}$ and $F_{PB}$ are equal) then
$$
\frac{|AP|-|AP|_0}{|BP|-|BP|_0}=\frac{\sin\theta}{\cos\theta}\frac{|AP|_0}{|BP|_0}
$$
or as expected
$$
\frac{4 \cos \theta-3}{4 \sin\theta-1} = 3\left(\frac{\sin\theta}{\cos\theta}\right)
$$
A: Well, here's my other attempt which does not assume that P moves down vertically:
I'll be using Hooke's law $T=\frac{\lambda x}{a}$
Let h = the height of the triangle in the diagram, i.e. the distance through which P falls.
$h=L_2\,cos\,\theta=L_1\,sin\,\theta$
and so
$L_1=\frac{L_2\,cos\,\theta}{sin\,\theta}$ - call this "Equation A"
$L_2=\frac{L_1\,sin\,\theta}{cos\,\theta}$ - call this "Equation B"
Resolving forces horizontally:
$T_1\,cos\,\theta-T_2\,sin\,\theta=0$ - Call this "Equation 1"
and Hooke's law gives us:
$T_1=\frac{\lambda(L_1-\frac{3}{20})}{\frac{3}{20}}$
and
$T_2=\frac{\lambda(L_2-\frac{1}{20})}{\frac{1}{20}}$
And so Eqaution 1 gives us:
$\frac{20}{3}\lambda(L_1-\frac{3}{20})\,cos\,\theta-20\lambda(L_2-\frac{1}{20})\,sin\,\theta=0$
Substituting fot $L_1$ from Equation A gives:
$\lambda(\frac{20}{3}\times\frac{L_2\,cos\,\theta}{sin\,\theta}-1)\,cos\,\theta=\lambda(20L_2-1)\,sin\,\theta$
Which rearranges to:
$L_2=\frac{3\,sin\theta(cos\,\theta-sin\,\theta)}{20\,cos^2\theta-60\,sin^2\theta}$
Similarly substituting for $L_2$ from Equation B leads to:
$L_1=\frac{3\,cos\theta(cos\,\theta-sin\,\theta)}{20\,cos^2\theta-60\,sin^2\theta}$
Now we are asked to find $\frac{L_1 - \frac{3}{20}}{L_2-\frac{1}{20}}$
Which is equal to:
$\frac{60\,cos\,\theta(cos\,\theta-sin\,\theta)-3(20\,cos^2\theta-60\,sin^2\theta)}{20(cos^2\theta-60\,sin^2\theta)}/\frac{60\,sin\theta(cos\,\theta-sin\,\theta)-(20\,cos^2\theta-60\,sin^2\theta)}{20(cos^2\theta-60\,sin^2\theta)}$
Which simplifies to:
$\frac{9\,sin^2\theta-3\,sin\,\theta\,cos\,\theta}{3\,sin\,\theta\,cos\,\theta-cos^2\theta}$
Which is not the required answer.
