Pentagon, find central angle?

Hi Dragan! Are we having drawing problems today? ;-)

Let O be the central point, let D be the topmost vertex,

and label the remaining vertices clockwise: E, F, G, H.

Let ∠HDO = B = ∠FOE and let ∠EDO = A = ∠HOG.

Then ∠X = ∠HDE = ∠GOG = A+B (so the ∠s about O sum to 360)

=> ΔHDE ≅ ΔFOG (by SAS) => HE = GF = c.

Hence ΔHOE is equilateral and ∠HOE = 60 = 2(A+B)

Thus, your ∠X = 30

in case you draw a circle around the pentagon and the two the pentagon and circle share an identical center then you definately see why that's seventy two because of the fact the circle and the pentagon could be divided in 5 equivalent areas from the middle. for this reason the middle attitude for the two a type of is seventy two tiers (360/5=seventy two).

From your given DIAGRAM, the angle X is opposit to side c,

and is in between sides a and b.

Thus

cos(X) = (a^2 + b^2 - c^2) / 2ab

<X = arccos[(a^2 + b^2 - c^2) / 2ab]

............... A N S W E R ...................

From the picture it is clear that all the angles at the center are different. Hence according to the given conditions --

Cos x = [ a² + b ² - c² ] / 2 ab

=> X = Arc Cos [ a² + b ² - c² ] / 2 ab ............... Answer

'''''''''''''''''''''''''''''''''''''''

a=b=c

X=60'

60+60+60=180'

360 degrees/5 sides = 72 degrees ---> final answer!