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°.

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

Hence, ∠BCD = 120°.

(iii) Join FC.

We know that,

Angle in a semicircle is a right angle.

∴ ∠FBC = 90°

In triangle BCD,

⇒ BC = CD (Sides of regular hexagon)

⇒ ∠CBD = ∠CDB = x (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠BCD + ∠CDB + ∠CBD = 180°

⇒ 120° + x + x = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = $\dfrac{60°}{2}$ = 30°.

From figure,

⇒ ∠FBD = ∠FBC - ∠CBD = 90° - x = 90° - 30° = 60°.

We know that,

The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.

⇒ ∠PDE = ∠FBD = 60°.

Hence, ∠PDE = 60°.