The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon.

Find the angles of triangle ABC.

Join OA, OB and OC.

Since, AB is the side of regular pentagon,

∠AOB = $\dfrac{360°}{5}$ = 72°.

Since, AC is the side of regular hexagon,

∠AOC = $\dfrac{360°}{6}$ = 60°.

From figure,

⇒ ∠AOB + ∠AOC + reflex∠BOC = 360°

⇒ 72° + 60° + reflex∠BOC = 360°

⇒ reflex∠BOC + 132° = 360°

⇒ reflex∠BOC = 360° - 132° = 228°.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

∴ ∠BOC = 2∠BAC

⇒ ∠BAC = $\dfrac{1}{2}∠BOC = \dfrac{1}{2} \times 228°$ = 114°.

Arc AC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

∴ ∠AOC = 2∠ABC

⇒ ∠ABC = $\dfrac{1}{2} \times ∠AOC = \dfrac{1}{2} \times 60°$ = 30°.

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = $\dfrac{1}{2} \times ∠AOB = \dfrac{1}{2} \times 72°$ = 36°.

Hence, angles of triangle are ∠ABC = 30°, ∠ACB = 36° and ∠BAC = 114°.