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 :

1. ∠ADC,
2. ∠BDA,
3. ∠ABC,
4. ∠AEC.

Answer 1

Join BC, BO, CO and EO

Since BD is the side of a regular hexagon,

`∠BOD = 360^circ/6 = 60^circ`

Since DC is the side of a regular pentagon,

`∠COD = 360^circ/5 = 72^circ`

In ∆BOD, ∠BOD = 60° and OB = OD

∴ ∠OBD = ∠ODB = 60°

i. In ∆OCD, ∠COD = 72° and OC = OD

∴ `∠ODC = 1/2 (180^circ - 72^circ)`

= `1/2 xx 108^circ`

= 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^circ - 72^circ)`

= `1/2 xx 108^circ`

= 54°

iv. In cyclic quadrilateral AECD

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

`=>` ∠AEC + 54° = 180°

`=>` ∠AEC = 180° – 54°

`=>` ∠AEC = 126°