In the given diagram, ABCDEF is a regular hexagon inscribed in a circle with centre O. PQ is a tangent to the circle at D. Find the value of :

(a) ∠FAG

(b) ∠BCD

(c) ∠PDE

(a) By formula,

Each interior angle of a n sided polygon = $\dfrac{(n - 2) \times 180°}{n}$

$= \dfrac{(6 - 2) \times 180°}{6} \\[1em] = \dfrac{4 \times 180°}{6} \\[1em] = \dfrac{2}{3} \times 180° \\[1em] = 120°.$

∴ ∠FAB = 120°

From figure,

∠FAG and ∠FAB form a linear pair.

∴ ∠FAB + ∠FAG = 180°

⇒ 120° + ∠FAG = 180°

⇒ ∠FAG = 180° - 120° = 60°.

Hence, ∠FAG = 60°.

(b) Each interior angle of regular hexagon = 120°.

Hence, ∠BCD = 120°.

(c) In triangle DEF,

⇒ ∠DEF = 120° (Each interior angle of a regular hexagon equals to 120°)

⇒ DE = EF (Sides of a regular hexagon)

⇒ ∠EDF = ∠EFD = a (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠DEF + ∠EDF + ∠EFD = 180°

⇒ 120° + a + a = 180°

⇒ 2a = 180° - 120°

⇒ 2a = 60°

⇒ a = $\dfrac{60}{2}$ = 30°

⇒ ∠EFD = 30°.

By alternate segment theorem,

The angle between a tangent to a circle and a chord drawn from the point of contact is equal to the angle subtended by that chord in the alternate segment of the circle.

∴ ∠PDE = ∠EFD = 30°.

Hence, ∠PDE = 30°.