Question

 If A, B, P and Q lie on the circle, centre O and $\angle APB={{56}^{0}}$ in the following figure:

Find the value of $x$ and $y$ .

Answer 1

Hint: For solving this question first we will prove “Inscribed Angle Theorem” and “Angle in the Same Segment Theorem” of the circle. After that, we will use them to find the answer to the question easily and ultimately we will find the value of $x$ and $y$ .

Complete Step-by-Step solution:
Given:
It is given that points A, B, P and Q lie on the circle, centre O and $\angle APB={{56}^{0}}$ in the following figure:

And we have to find the value of $x$ and $y$ .
Now, before we proceed we will see 2 important properties related to triangles one by one which will be used to prove two important theorems of circle.
First property:
It is also known as “Exterior Angle Theorem”. It states that if a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior angles. As shown in the figure below:

In the above figure, $\Delta ABC$ is shown in which side $BC$ is extended to $D$ . Then, from exterior angle theorem, we can write, $\angle DCA=\angle ABC+\angle CAB$ .
Second property:
Angles opposite to equal sides of an isosceles triangle are equal. This can be understood with the help of the figure below:

In the above figure, $\Delta ABC$ is shown in which $AB=AC$ . Then, $\angle ABC=\angle ACB$ .
Now, we will use the above properties collectively to prove two important theorems of circle.
Now, we should know the following two important theorems of circle:
Theorem 1: Inscribed Angle Theorem
Statement: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Proof: Consider there is an arc PQ of a circle subtending $\angle POQ$ at the centre O and $\angle PAQ$ at a point A on the remaining part of the circle. We need to prove that $\angle POQ=2\angle PAQ$ . For more clarity look at the figure given below:
 

In the above figure, we have joined point A and O and extended it to a point B.
Now, as we know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles. Then,
$\begin{align}
  & \angle QOB=\angle QAO+\angle OQA............\left( 1 \right) \\
 & \angle POB=\angle PAO+\angle OPA..............\left( 2 \right) \\
\end{align}$
Also, in $\Delta OAQ$ and $\Delta OPA$ :
$\begin{align}
  & OA=OQ=\text{ radius of the circle} \\
 & OA=OP=\text{ radius of the circle} \\
\end{align}$
Thus, $\Delta OAQ$ and $\Delta OPA$ will be isosceles triangles. And as we know that angles opposite to equal sides of an isosceles triangle are equal. Then,
$\begin{align}
  & \angle QAO=\angle OQA \\
 & \angle PAO=\angle OPA \\
\end{align}$
Now, in equation (1) put $\angle OQA=\angle QAO$ and in equation (2) put $\angle OPA=\angle PAO$ . Then,
$\begin{align}
  & \angle QOB=\angle QAO+\angle OQA \\
 & \Rightarrow \angle QOB=2\angle QAO.............\left( 3 \right) \\
 & \angle POB=\angle PAO+\angle OPA \\
 & \Rightarrow \angle POB=2\angle PAO.............\left( 4 \right) \\
\end{align}$
Now, add equation (3) and (4). Then,
$\begin{align}
  & \angle QOB+\angle POB=2\angle QAO+2\angle PAO \\
 & \Rightarrow \angle POQ=2\left( \angle QAO+\angle PAO \right) \\
 & \Rightarrow \angle POQ=2\angle PAQ \\
\end{align}$
Hence, proved.
Theorem 2: Angle in the Same Segment Theorem
Statement: Angles which are in the same segment are equal, i.e. angles subtended by the same arc at the circumference are equal.
Proof:
In the previous theorem if we join points P and Q and form a chord PQ in the above figure. Then, $\angle PAQ$ is also called the angle formed in the same segment PAQP. For more clarity look at the figure given below:

Now, as in the previous theorem, A can be any point on the remaining circle. So, if we take any other point C on the remaining part of the circle. Then,
$\begin{align}
  & \angle POQ=2\angle PCQ=2\angle PAQ \\
 & \Rightarrow \angle PCQ=\angle PAQ \\
\end{align}$
Therefore, this proves that angles in the same segment of a circle are equal.
Now, we will be using the above two theorems to solve this question.
As we have the following figure:

Now, from the first theorem, we conclude that $\angle AOB=2\angle APB$ and from the second theorem, we conclude that $\angle AQB=\angle APB$ . Then,
\[\begin{align}
  & \angle AOB=2\angle APB \\
 & \Rightarrow {{x}^{0}}=2\times {{56}^{0}} \\
 & \Rightarrow {{x}^{0}}={{112}^{0}} \\
 & \angle AQB=\angle APB \\
 & \Rightarrow {{y}^{0}}={{56}^{0}} \\
\end{align}\]
Now, from the above result, we conclude that the value of $x={{112}^{0}}$ and $y={{56}^{0}}$.
Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, though the question is very easy, we should apply the results “Inscribed Angle Theorem” and “Angle in the Same Segment Theorem” with full clarity, so that we can find the correct answer easily.