#### Question

In the given figure, BD is a side of a regular hexagon, DC is a side of a regular pentagon and AD is a diameter. Calculate :

(i) ∠ADC (ii) ∠BDA, (iii) ∠ABC, (iv) ∠AEC.

#### Solution

Join BC, BO, CO and EO

Since BD is the side of a regular hexagon,

`∠BOD = = 360/6 =60°`

Since DC is the side of a regular pentagon,

`∠COD = 360/5 = 72°`

In ∆BOD. ∠BOD = 60° and OB = OD

∴ ∠OBD = ∠ODB = 60°

(i) In ∆OCD, ∠COD = 72° and OC = OD

∴ ∠ODC =` 1/2 (180° - 72°) `

= `1/2 xx 108°`

= 54°

Or, ∠ADC = 54°

(ii) ∠BDO = 60° or ∠BDA = 60°

(iii) Arc AC subtends ∠AOC at the centre and

∠ABC at the remaining part of the circle.

∴ `∠ ABC = 1 /2∠AOC`

= `1/2[∠AOD - ∠COD]`

= `1/2 xx (180° - 72°) `

= `1/2 xx108°`

= 54°

(iv) In cyclic quadrilateral AECD

∠AEC + ∠ADC = 180° [sum of opposite angles]

⇒ ∠AEC + 54° = 180°

⇒ ∠AEC = 180° - 54°

⇒ ∠AEC = 126°