Rotating the Pentagon

Is there a way to solve it without a calculator?

My logic:

Let a be the smaller part of the bigger side.

Let b the bigger part of the bigger side.

Let c be the smaller side.

Answer = $\frac{a+b}{c}$

Note that the internal angle of a regular pentagon is $\frac{(n-2)\cdot{}180^{\circ{}}}{n}=\frac{(3)\cdot{}180^{\circ{}}}{5}=108^{\circ{}}$

Then apply the law of sines.

$\frac{\sin{(20^{\circ{}})}}{a}=\frac{\sin{(52^{\circ{}})}}{b}=\frac{\sin{(108^{\circ{}})}}{c}$

Rearranging:

$\frac{\frac{\sin{(20^{\circ{}})}\cdot{}(c)}{\sin{(108^{\circ{}})}}+\frac{\sin{(52^{\circ{}})}\cdot{}(c)}{\sin{(108^{\circ{}})}}}{c}=\frac{\sin{(20^{\circ{}})}}{\sin{(108^{\circ{}})}}+\frac{\sin{(52^{\circ{}})}}{\sin{(108^{\circ{}})}}$

$\cong{}1.1881848003464956036128466432226444493228930328779325$