How do you find the measures of the numbered angle in each kite below?

Hint: We will use the property of isosceles and right-angle triangles to solve the above question. To find angles 1 and 2 in figure 1 we will at first join angles 1 and 2 and then we will use the property that if two sides of the triangle is equal then their opposite angle is also equal. And, in figure 2 we will first solve the right angle triangle and find angle 2 and then we will find angle 1.


Complete step by step answer:
We will use the first recall of the property of the isosceles triangle to solve the above question. Since we know that in an isosceles triangle two sides are equal and their corresponding is also equal.
So, at first, we will solve for figure 1.
Let A, B, C, D denotes the four vertices of the kite then:



We will first join vertices BC. From the figure we can see that $ \angle BAC=120{}^\circ $ , $ \angle BDC=44{}^\circ $ , and side AB = AC and Side BD = CD
So, in $ \Delta ABC $ we have side AB = BC, so $ \Delta ABC $ is an isosceles triangle. So, we can say that $ \angle ABC=\angle ACB $ (Property of isosceles triangle)
We know that sum of all angle of a triangle is equal to $ 180{}^\circ $.
 $ \Rightarrow \angle ABC+\angle ACB+\angle BAC=180{}^\circ $
Since, we know that $ \angle BAC=120{}^\circ $ and $ \angle ABC=\angle ACB $ .
 $ \begin{align}
  & \Rightarrow \angle ABC+\angle ABC+120{}^\circ =180{}^\circ \\
 & \Rightarrow 2\angle ABC=60{}^\circ \\
 & \therefore \angle ABC=30{}^\circ \\
\end{align} $
So, $ \angle ABC=\angle ACB=30{}^\circ ---\left( 1 \right) $
So, in $ \Delta DBC $ we have side DB = DC, so $ \Delta DBC $ is an isosceles triangle. So, we can say that $ \angle DBC=\angle DCB $ (Property of isosceles triangle)
We know that sum of all angle of triangle is equal to $ 180{}^\circ $ .
 $ \Rightarrow \angle DBC+\angle DCB+\angle BDC=180{}^\circ $
Since, we know that $ \angle BDC=44{}^\circ $ and $ \angle DBC=\angle DCB $ .
\[\begin{align}
  & \Rightarrow \angle DBC+\angle DBC+44{}^\circ =180{}^\circ \\
 & \Rightarrow 2\angle ABC=136{}^\circ \\
 & \therefore \angle ABC=68{}^\circ \\
\end{align}\]
So, $ \angle ABC=\angle ACB=68{}^\circ ---\left( 2 \right) $
Now, from figure we can see that $ \angle 1=\angle ABD=\angle ABC+\angle DBC $
 $ \begin{align}
  & \Rightarrow \angle 1=\angle ABD=30{}^\circ +68{}^\circ \\
 & \therefore \angle 1=98{}^\circ \\
\end{align} $ (From (1) and (2))
Similarly, from figure we can write $ \angle 2=\angle ACD=\angle ACB+\angle DCB $
 $ \begin{align}
  & \Rightarrow \angle 2=\angle ACD=30{}^\circ +68{}^\circ \\
 & \therefore \angle 2=98{}^\circ \\
\end{align} $ (From (1) and (2))

Now, we will solve for figure 2.
Let A, B, C, D denotes the four vertices of the kite then:



We will at first prove the congruency of the triangle $ \Delta ABC $ and $ \Delta ACB $ :
AB = AC (From figure)
CD = BD (From figure)
AD = AD (common)
So, $ \Delta ABC\tilde{=}\Delta ACB $
Thus, $ \angle BDE=\angle CDE---\left( 3 \right) $
Now, in $ \Delta BDC $ , we have BD = DC. So, $ \Delta BDC $ is an isosceles triangle.
Therefore, $ \angle DBC=\angle DCB $
We know that sum of all angle of a triangle is equal to $ 180{}^\circ $.
 $ \Rightarrow \angle BDC+\angle DBC+\angle DCB=180{}^\circ $
Since, we know that $ \angle DCB=70{}^\circ $ and $ \angle DBC=\angle DCB=70{}^\circ $ .
\[\begin{align}
  & \Rightarrow 70{}^\circ +\angle BDC+70{}^\circ =180{}^\circ \\
 & \Rightarrow \angle BDC=180{}^\circ -140{}^\circ \\
 & \therefore \angle BDC=40{}^\circ \\
\end{align}\]
So, $ \angle BDC=40{}^\circ $
From, from figure we can see from figure that $ \angle BDC=\angle BDE+\angle CDE $
And, from (3) we have $ \angle BDE=\angle CDE $
We can also see from figure that $ \angle CDE=\angle 2 $
 $ \begin{align}
  & \Rightarrow \angle BDC=\angle CDE+\angle CDE \\
 & \Rightarrow \angle BDC=2\angle CDE \\
 & \Rightarrow \angle BDC=2\angle 2 \\
 & \Rightarrow 40{}^\circ =2\angle 2 \\
 & \therefore \angle 2=20{}^\circ \\
\end{align} $
Now, in triangle $ \Delta DCE $ :
 $ \angle DEC+\angle DCE+\angle EDC=180{}^\circ $
And, we know that $ \angle CDE=20{}^\circ $ and $ \angle DCE=70{}^\circ $ .
  $ \begin{align}
  & \Rightarrow \angle DEC+20{}^\circ +70{}^\circ =180{}^\circ \\
 & \Rightarrow \angle DEC=90{}^\circ \\
\end{align} $
Now, we can see from figure 2 that $ \angle CED=\angle AEB $ because they are vertically opposite angle:
 $ \Rightarrow \angle CED=\angle AEB=90{}^\circ $
We can also see figure 2 that $ \angle AEB=\angle 2 $ .
So, $ \angle 2=90{}^\circ $
 This is our required solution.

Note:
 We should first recall the property of isosceles triangle and they are required to note that diagonal of the kite always intersect each other at $ 90{}^\circ $. So, we can also use this property of kite to solve the above question.