The diagram above shows a kite ABCD where angle BAC = 30°. EB, BD and DF are three sides of a regular polygon. ABE and ADF are straight lines. The ratio of angle CBE to angle CBD is 3 : 2.

1. Find the interior angle of the regular polygon at $B$.

In the kite, $AC$ is the line of symmetry, so it bisects $\angle BAD$.

Given $\angle BAC = 30^\circ$,
$\angle CAD = 30^\circ$
Hence
$\angle BAD = 30^\circ + 30^\circ = 60^\circ.$
Since $AB = AD$, triangle $ABD$ is isosceles. Therefore
$\angle ABD = \angle BDA = \frac{180^\circ - 60^\circ}{2} = 60^\circ.$
Since $ABE$ is a straight line,
$\angle EBD = 180^\circ - \angle ABD = 180^\circ - 60^\circ = 120^\circ.$
So the interior angle of the regular polygon is $120^\circ$.

2. Use the interior angle formula to find the number of sides.

For a regular polygon with $n$ sides,
$\text{interior angle} = \frac{(n-2)180^\circ}{n}.$
So
$\frac{(n-2)180}{n} = 120$
$180n - 360 = 120n$
$60n = 360$
$n = 6.$

1. Find $\angle CBD$ using the hexagon interior angle and the given ratio.

From part (a), the polygon has $6$ sides, so each interior angle is $120^\circ$.

Since $EB$ and $BD$ are two sides of the regular polygon,
$\angle EBD = 120^\circ.$

Given
$\angle CBE : \angle CBD = 3:2,$
and $\angle CBE + \angle CBD = 120^\circ$,
$\angle CBD = \frac{2}{3+2}\times 120^\circ = 48^\circ.$

2. Use the kite property to find $\angle BCD$.

In kite $ABCD$, $CB = CD$, so triangle $BCD$ is isosceles.

Hence
$\angle CBD = \angle BDC = 48^\circ.$

Therefore,
$\angle BCD = 180^\circ - 48^\circ - 48^\circ = 84^\circ.$