Star in a regular pentagon

Let us consider one-fifth of the pentagon. Let the center of the pentagon by $O$ , the side of the pentagon be $a$ , two adjacent vertices of the pentagon be $A$ and $B$ , and the other vertex of the non-shaded triangle be $P$ .

We note that $\angle OAB = 54^\circ$ and $\angle PAB = 36^\circ$ . And the area of $\triangle OAB = \frac {1}{5}$ of the area of pentagon is given by:

$\quad A_{OAB} = \frac {1}{2} a (\frac {1}{2} a \tan {54^\circ})$

Similarly, he area of $\triangle PAB = \frac {1}{5}$ of total non-shaded area is:

$\quad A_{PAB} = \frac {1}{2} a (\frac {1}{2} a \tan {36^\circ})$

Therefore, the portion of non-shaded area is:

$\quad \dfrac {A_{OAB}}{A_{PAB}} = \dfrac {\tan {36^\circ}}{\tan {54^\circ}}$

The portion the star occupies:

$\quad 1- \dfrac {A_{OAB}}{A_{PAB}} = 1 - \dfrac {\tan {36^\circ}}{\tan {54^\circ}} = 1 - \dfrac {0.726542528}{1.37638192}$

$\quad \quad \quad \quad \quad \quad = 0.472135955 \approx \boxed {47.2} \%$

| $\text{AB is the side of the pentagon. M its midpoint. O the center}\\\text{and AO=r.}\\right~\angle d ~\Delta~AMO =\dfrac 1 {10} ~of~the~ Polygon.\\\Delta~AMC=\dfrac 1 {10}~ of~ unshaded ~area.\\\text{The angles in the sketch are from pentagon properties.}\\\text{Sides are from trigonometry.}\\Area~\Delta ~AMC= \dfrac 1 2 *rCos54^o*(rCos54^o * Tan36^o) \\Area~\Delta ~AMO=\dfrac 1 2 *rCos54^o*rSin(18^o+36^o).\\ Fraction~~of~star~=1- \dfrac{ Area~\Delta ~AMC}{Area~\Delta ~AMO}\\=\dfrac{\dfrac 1 2 *rCos54^o*r\color{#D61F06}{(Cos54^o*Tan36^o)} }{\dfrac 1 2 *rCos54^o*r\color{#D61F06}{Sin54^o} } \\=1-\dfrac{Tan36^o}{Cot54^o }=1-(Tan36^o)^2\\\%= \Huge 47.2$

Am I suppose to use the calculator??