geometry
For reference:
In the figure two regular pentagons are shown. Calculate "x".
My progress..I marked the angles I found but I couldn't find the equation to solve
$a_i = \frac{180(5-2)}{5}=108^o\\ \triangle BCG (isosceles) \therefore \measuredangle CGB = 36^o$
If we find $\measuredangle DCF$ the problem is solved because $\measuredangle DJF$ is half of $\measuredangle DCF$
$\triangle ABC$ is isosceles. Therefore,
$\measuredangle BAC = \measuredangle C \implies 2\alpha+\alpha =120^\circ\implies \alpha =40^\circ$
Therefore $\measuredangle ABC = 180^\circ - 160^\circ = 20^\circ$
Draw $AX$ such that $\measuredangle XAK = 30^\circ$.
Draw $TX$.
$\measuredangle CDB = 120^\circ ~~\therefore \measuredangle DIA = 90^\circ$
Since $CI$ is angle bisector $IA = IX$
$\triangle ITA \cong \triangle IXT ~(L.A.L) \rightarrow \measuredangle AXT = 60^\circ$
Therefore $\triangle ATX$ is equilateral.
$\measuredangle KXB = 180 ^\circ -110^\circ= 70^\circ$
$AK$ is angle bisector and the height of triangle $ATX$
$ \therefore AKT = 90^\circ$
$TK = KX \rightarrow \triangle KTB \cong \triangle KXB~ (L.A.L)$
Therefore $\triangle TBX$ is isosceles.
$\measuredangle B = 180^\circ -140^\circ =40^\circ$
$\therefore \boxed{\color{red}x=40^\circ-20^\circ=20^\circ}$
$$45^{\circ}=\measuredangle EMC=\measuredangle EBC+\measuredangle ACB=\frac{1}{2}\left(\widehat{EC}+\widehat{AB}\right)=\frac{3}{2}\widehat{AB}=\frac{3}{2}\cdot\frac{360^{\circ}}{n}$$
Best Answer
$\angle{FDC} = \angle{DFC} = 180 - 108 = 72$
Therefore, $\angle{DCF} = 180 - 72 \times 2 = 36$
Therefore, $\angle{DCG} = 36 + 108 = 144$
Therefore, $\angle{CDG} = \angle{CGD} = \frac{180 - 144}2 = 18$
Therefore, $\angle{JDF} = 180 - 108 -18 = 54$
Therefore, $x = \angle{DJF} = 180 - 54 \times 2 = 72$.