Question

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

Answer 1

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 $\mathrm{\angle }BAC=120{}^{\circ }$ , $\mathrm{\angle }BDC=44{}^{\circ }$ , and side AB = AC and Side BD = CD
So, in $\mathrm{\Delta }ABC$ we have side AB = BC, so $\mathrm{\Delta }ABC$ is an isosceles triangle. So, we can say that $\mathrm{\angle }ABC=\mathrm{\angle }ACB$ (Property of isosceles triangle)
We know that sum of all angle of a triangle is equal to $180{}^{\circ }$.
 $\Rightarrow \mathrm{\angle }ABC+\mathrm{\angle }ACB+\mathrm{\angle }BAC=180{}^{\circ }$
Since, we know that $\mathrm{\angle }BAC=120{}^{\circ }$ and $\mathrm{\angle }ABC=\mathrm{\angle }ACB$ .
 $\begin{array}{rl}& \Rightarrow \mathrm{\angle }ABC+\mathrm{\angle }ABC+120{}^{\circ }=180{}^{\circ }\\ & \Rightarrow 2\mathrm{\angle }ABC=60{}^{\circ }\\ & \therefore \mathrm{\angle }ABC=30{}^{\circ }\end{array}$
So, $\mathrm{\angle }ABC=\mathrm{\angle }ACB=30{}^{\circ }---\left(1\right)$
So, in $\mathrm{\Delta }DBC$ we have side DB = DC, so $\mathrm{\Delta }DBC$ is an isosceles triangle. So, we can say that $\mathrm{\angle }DBC=\mathrm{\angle }DCB$ (Property of isosceles triangle)
We know that sum of all angle of triangle is equal to $180{}^{\circ }$ .
 $\Rightarrow \mathrm{\angle }DBC+\mathrm{\angle }DCB+\mathrm{\angle }BDC=180{}^{\circ }$
Since, we know that $\mathrm{\angle }BDC=44{}^{\circ }$ and $\mathrm{\angle }DBC=\mathrm{\angle }DCB$ .
$$\begin{array}{rl}& \Rightarrow \mathrm{\angle }DBC+\mathrm{\angle }DBC+44{}^{\circ }=180{}^{\circ }\\ & \Rightarrow 2\mathrm{\angle }ABC=136{}^{\circ }\\ & \therefore \mathrm{\angle }ABC=68{}^{\circ }\end{array}$$
So, $\mathrm{\angle }ABC=\mathrm{\angle }ACB=68{}^{\circ }---\left(2\right)$
Now, from figure we can see that $\mathrm{\angle }1=\mathrm{\angle }ABD=\mathrm{\angle }ABC+\mathrm{\angle }DBC$
 $\begin{array}{rl}& \Rightarrow \mathrm{\angle }1=\mathrm{\angle }ABD=30{}^{\circ }+68{}^{\circ }\\ & \therefore \mathrm{\angle }1=98{}^{\circ }\end{array}$ (From (1) and (2))
Similarly, from figure we can write $\mathrm{\angle }2=\mathrm{\angle }ACD=\mathrm{\angle }ACB+\mathrm{\angle }DCB$
 $\begin{array}{rl}& \Rightarrow \mathrm{\angle }2=\mathrm{\angle }ACD=30{}^{\circ }+68{}^{\circ }\\ & \therefore \mathrm{\angle }2=98{}^{\circ }\end{array}$ (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 $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }ACB$ :
AB = AC (From figure)
CD = BD (From figure)
AD = AD (common)
So, $\mathrm{\Delta }ABC\widetilde{=}\mathrm{\Delta }ACB$
Thus, $\mathrm{\angle }BDE=\mathrm{\angle }CDE---\left(3\right)$
Now, in $\mathrm{\Delta }BDC$ , we have BD = DC. So, $\mathrm{\Delta }BDC$ is an isosceles triangle.
Therefore, $\mathrm{\angle }DBC=\mathrm{\angle }DCB$
We know that sum of all angle of a triangle is equal to $180{}^{\circ }$.
 $\Rightarrow \mathrm{\angle }BDC+\mathrm{\angle }DBC+\mathrm{\angle }DCB=180{}^{\circ }$
Since, we know that $\mathrm{\angle }DCB=70{}^{\circ }$ and $\mathrm{\angle }DBC=\mathrm{\angle }DCB=70{}^{\circ }$ .
$$\begin{array}{rl}& \Rightarrow 70{}^{\circ }+\mathrm{\angle }BDC+70{}^{\circ }=180{}^{\circ }\\ & \Rightarrow \mathrm{\angle }BDC=180{}^{\circ }-140{}^{\circ }\\ & \therefore \mathrm{\angle }BDC=40{}^{\circ }\end{array}$$
So, $\mathrm{\angle }BDC=40{}^{\circ }$
From, from figure we can see from figure that $\mathrm{\angle }BDC=\mathrm{\angle }BDE+\mathrm{\angle }CDE$
And, from (3) we have $\mathrm{\angle }BDE=\mathrm{\angle }CDE$
We can also see from figure that $\mathrm{\angle }CDE=\mathrm{\angle }2$
 $\begin{array}{rl}& \Rightarrow \mathrm{\angle }BDC=\mathrm{\angle }CDE+\mathrm{\angle }CDE\\ & \Rightarrow \mathrm{\angle }BDC=2\mathrm{\angle }CDE\\ & \Rightarrow \mathrm{\angle }BDC=2\mathrm{\angle }2\\ & \Rightarrow 40{}^{\circ }=2\mathrm{\angle }2\\ & \therefore \mathrm{\angle }2=20{}^{\circ }\end{array}$
Now, in triangle $\mathrm{\Delta }DCE$ :
 $\mathrm{\angle }DEC+\mathrm{\angle }DCE+\mathrm{\angle }EDC=180{}^{\circ }$
And, we know that $\mathrm{\angle }CDE=20{}^{\circ }$ and $\mathrm{\angle }DCE=70{}^{\circ }$ .
  $\begin{array}{rl}& \Rightarrow \mathrm{\angle }DEC+20{}^{\circ }+70{}^{\circ }=180{}^{\circ }\\ & \Rightarrow \mathrm{\angle }DEC=90{}^{\circ }\end{array}$
Now, we can see from figure 2 that $\mathrm{\angle }CED=\mathrm{\angle }AEB$ because they are vertically opposite angle:
 $\Rightarrow \mathrm{\angle }CED=\mathrm{\angle }AEB=90{}^{\circ }$
We can also see figure 2 that $\mathrm{\angle }AEB=\mathrm{\angle }2$ .
So, $\mathrm{\angle }2=90{}^{\circ }$
 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.