Key Concept: Kite properties, Angle relationships
b) $100^\circ$
[Solution Description]
First, recall that in a kite, one diagonal bisects the angles at the vertices it connects. So, diagonal $BD$ bisects $\angle ABC$ and $\angle ADC$.
Given $\angle ABC = 120^\circ$, so $\angle ABO = \angle CBO = \frac{120^\circ}{2} = 60^\circ$.
Similarly, $\angle ADC = 80^\circ$, so $\angle ADO = \angle CDO = \frac{80^\circ}{2} = 40^\circ$.
Now, consider triangle $AOD$. The sum of angles in a triangle is $180^\circ$.
We know $\angle OAD$ can be found using the property that diagonals of a kite are perpendicular. Thus, $\angle AOD = 90^\circ$.
However, let's verify this by calculating $\angle OAD$:
In quadrilateral $ABCD$, the sum of all angles is $360^\circ$.
So, $\angle BAD + \angle ABC + \angle BCD + \angle CDA = 360^\circ$.
Since $AB = BC$ and $CD = DA$, we can find $\angle BAD = \angle BCD$ as follows:
Let $\angle BAD = x$, then $\angle BCD = x$ (because $AB = BC$ and $CD = DA$).
So, $x + 120^\circ + x + 80^\circ = 360^\circ$.
Which gives $2x + 200^\circ = 360^\circ$.
So, $2x = 160^\circ$ and $x = 80^\circ$.
Now, $\angle OAD = \frac{\angle BAD}{2} = \frac{80^\circ}{2} = 40^\circ$.
In triangle $AOD$, $\angle AOD + \angle OAD + \angle ODA = 180^\circ$.
Substituting known values: $\angle AOD + 40^\circ + 40^\circ = 180^\circ$.
So, $\angle AOD = 100^\circ$.
Your Answer is correct.
b) $100^\circ$
[Solution Description]
First, recall that in a kite, one diagonal bisects the angles at the vertices it connects. So, diagonal $BD$ bisects $\angle ABC$ and $\angle ADC$.
Given $\angle ABC = 120^\circ$, so $\angle ABO = \angle CBO = \frac{120^\circ}{2} = 60^\circ$.
Similarly, $\angle ADC = 80^\circ$, so $\angle ADO = \angle CDO = \frac{80^\circ}{2} = 40^\circ$.
Now, consider triangle $AOD$. The sum of angles in a triangle is $180^\circ$.
We know $\angle OAD$ can be found using the property that diagonals of a kite are perpendicular. Thus, $\angle AOD = 90^\circ$.
However, let's verify this by calculating $\angle OAD$:
In quadrilateral $ABCD$, the sum of all angles is $360^\circ$.
So, $\angle BAD + \angle ABC + \angle BCD + \angle CDA = 360^\circ$.
Since $AB = BC$ and $CD = DA$, we can find $\angle BAD = \angle BCD$ as follows:
Let $\angle BAD = x$, then $\angle BCD = x$ (because $AB = BC$ and $CD = DA$).
So, $x + 120^\circ + x + 80^\circ = 360^\circ$.
Which gives $2x + 200^\circ = 360^\circ$.
So, $2x = 160^\circ$ and $x = 80^\circ$.
Now, $\angle OAD = \frac{\angle BAD}{2} = \frac{80^\circ}{2} = 40^\circ$.
In triangle $AOD$, $\angle AOD + \angle OAD + \angle ODA = 180^\circ$.
Substituting known values: $\angle AOD + 40^\circ + 40^\circ = 180^\circ$.
So, $\angle AOD = 100^\circ$.