In the picture, the tangent to the circumcircle of a regular pentagon through a vertex is shown.

Calculate the angle when the tangent makes with the two sides of the pentagon through the point of contact.

Let us first label the diagram.

To find: the angle when the tangent makes with the two sides of the pentagon through the point of contact i.e. ∠EAX and ∠BAY

Construction: Join AD and BD

In ΔAED

AE = DE [Sides of regular polygon are equal]

∠EAD = ∠EDA [Angles opposite to equal sides are equal]

Also, In ΔAEC, By angle sum property

∠AED + ∠EAD + ∠EDA = 180°

⇒ 108° + ∠EDA + ∠EDA = 180° [∠AED = 108°, Each angle in regular pentagon is 108°]

⇒ 2 ∠EDA = 72°

⇒ ∠EDA = 36° …[1]

Similarly,

∠BDC = 36° …[2]

Also,

∠CDE = 108° [Each angle in regular pentagon is 108°]

⇒ ∠EDA + ∠ADB + ∠BDC = 108°

⇒ 36° + ∠ADB + 36° = 108° [From 1 and 2]

⇒ ∠ADB = 36°

Also, AB is a chord

⇒ ∠BAY = ∠ADB [By alternate segment theorem i.e. angle between chord and tangent is equal to the angle in the other segment]

⇒ ∠BAY = 36°

And By symmetry,

∠EAX = ∠BAY = 36°