What's that angle?

The colored dots indicate angles of equal measure. Due to symmetry $\triangle BCF$ is isosceles and is similar to $\triangle ADE$ . Therefore $\angle BFC = \angle AED = 108^\circ$ and $\angle CFD = 180^\circ - 108^\circ = \boxed{72^\circ}$ .

$BD$ is a rotation of $AC$ by an angle of $\dfrac{360^{\circ}}{5} = 72^{\circ}$ , so this is the angle between the two line segments.

You guys can do 1) Si=( n-3) * 180 and divide the result for 2 . so Si=540° where n= the number of sides of the figure that is 5.

2) the figure has equal sides so the diagonal makes 2 oposite angules that has the same value in the figures that appers as a parallelogram. So if we know 1 angle is 108° and the other angle near to the ? angle is 108° we also can conclude that 108° + ?° will give us 180 because the sum of an internal angle with the external angle is equal to 180°

3) then 108 + ?°= 180 ?°= 180-108 ?°=72

In a pentagon all angles add up to 540°. In a regular pentagon this means each of the 5 angles is 540°/5=108°. So angle A= angle B= angle C= angle D= angle E=108°.

Since in a regular pentagon all sides are of equal length that means that AB=BC=CD=DE=EA. So triangle ABC is an isosceles triangle with equal sides AB and BC and a top corner B of 108°. In an isosceles triangle both base angles are the same, in triangle ABC that means that angle A=angle C= $\frac{180°-108°}{2}$ =36°.

If we then look at triangle CDF we can see that we have an angle C of 108°-36°=72°. Angle D is the same angle as angle C in triangle ABC so angle D=36°. Angle F the angle we're looking for, then must be 180°-72°-36°=72°.

Is angle E is 108 degree and AEDF (F is that intersection point) is parallelogram so angleE=angleF now by linear theor. Required angle would be 180-108= 72.

In isosceles $\triangle BCD$ we see $\angle DBC=\angle BDC$ so it's clear that $\angle EDB = \angle ABD$ , by similarity of $\triangle FDC$ and $\triangle FAB$ , $\angle EDB = \angle ABD = \angle ACD$ , an angle in a regular pentagon is $108 = \angle EDB + \angle BDC = \angle ACD + \angle BDC$ , so in $\triangle BCD$ , $\angle ? = 180-108 = \boxed{72}$ .

Suppose the angle x. Observing the pantagon we can get the equation 180 - (x/2)*3 = 108 - 2/x . Solving this equation we get x = 72.