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.

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Solution

$BD$ is a side of a regular hexagon

(i) $\angle BOD = \frac{360}{6} = 60^o$

$DC$ is a side of regular pentagon

$\angle DOC = \frac{360}{5} = 72^o$

(ii) In $\triangle BOD, \angle BOD = 60^o$

$OB = OD$ (radius of the same circle)

$\therefore \angle OBD = \angle ODB = 60^o$

(iii) In $\triangle OCD, \angle COD = 72^o \ and \ OC = OD$

$\angle ODC = \frac{1}{2} (180-72) = 54^o$

(iv) In cyclic quadrilateral $AECD$

$\angle AEC + \angle ADC = 180^o$ (opposite angles of a cyclic quadrilateral are supplementary)

$\angle AEC = 180-54 = 126^o$

$\\$

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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.

is a side of a regular hexagon (i) is a side of regular pentagon (ii) In (radius of the same circle) (iii) In (iv) In cyclic quadrilateral (opposite angles of a cyclic quadrilateral are supplementary)

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.