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,

${\displaystyle \angle BOD=\frac{{360}^{\circ }}{6}={60}^{\circ }}$

Since DC is the side of a regular pentagon,

${\displaystyle \angle COD=\frac{{360}^{\circ }}{5}={72}^{\circ }}$

In ∆BOD, ∠BOD = 60° and OB = OD

∴ ∠OBD = ∠ODB = 60°

i. In ∆OCD, ∠COD = 72° and OC = OD

∴ ${\displaystyle \angle ODC=\frac{1}{2}({180}^{\circ }-{72}^{\circ })}$

= ${\displaystyle \frac{1}{2}\times {108}^{\circ }}$

= 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.

∴ ${\displaystyle \angle ABC=\frac{1}{2}\angle AOC}$

= ${\displaystyle \frac{1}{2}[\angle AOD-\angle COD]}$

= ${\displaystyle \frac{1}{2}\times ({180}^{\circ }-{72}^{\circ })}$

= ${\displaystyle \frac{1}{2}\times {108}^{\circ }}$

= 54°

iv. In cyclic quadrilateral AECD

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

${\displaystyle \Rightarrow }$ ∠AEC + 54° = 180°

${\displaystyle \Rightarrow }$ ∠AEC = 180° – 54°

${\displaystyle \Rightarrow }$ ∠AEC = 126°