Question

In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with center O.

Calculate the sizes of
i) $\angle AOB$
ii) $\angle ACB$
iii) $\angle ABC$

Answer 1

Hint: We solve this question by first finding the values of $\angle AOB$ and $\angle AOC$ using the property that angle subtended by any side of the regular n-sided polygon is equal to $\dfrac{2\pi }{n}$. Then we consider the chord AB and use the property, the angle subtended by any chord of the circle at the centre is double the angle subtended by the chord at any point on the circumference of the circle, to find the relation between $\angle AOB$ and $\angle ACB$. Then we substitute the value of $\angle AOB$ and find the value of $\angle ACB$. Then we consider the chord AC and use the same property to find the relation between $\angle AOC$ and $\angle ABC$. Then we substitute the value of $\angle AOC$ and find the value of $\angle ABC$.

Complete step-by-step solution:
First, let us remember the concept that the total angle made by the circle at the centre is $2\pi $.
We are given that AB is the side of a regular six-sided polygon.
In a regular n-sided polygon angle subtended by the line made by adjacent vertices, that is any side of the polygon is equal to $\dfrac{2\pi }{n}$.
Using this property, as AB is a side of a six-sided regular polygon, the angle subtended by any side at the centre can be given by
$\Rightarrow \dfrac{2\pi }{6}$

From the above diagram, we can see that angle subtended by the side AB is $\angle AOB$. So, we get
$\Rightarrow \angle AOB=\dfrac{2\pi }{6}=\dfrac{\pi }{3}.............\left( 1 \right)$
Similarly, we are given that AC is the side of a regular eight-sided regular polygon. So, the angle subtended by the side AC at the centre of circle O is,
$\Rightarrow \angle AOC=\dfrac{2\pi }{8}=\dfrac{\pi }{4}.............\left( 2 \right)$
Now let us consider the chord AB. We can see that the chord AB is subtending the angle $\angle AOB$ at the centre O and angle $\angle ACB$ at the point C.
Now let us consider another property.
The angle subtended by any chord of the circle at the centre is double the angle subtended by the chord at any point on the circumference of the circle.
Using this property, we can write that
$\Rightarrow \angle AOB=2\angle ACB$
Substituting the value of $\angle AOB$ form equation (1), we get
$\begin{align}
  & \Rightarrow \dfrac{\pi }{3}=2\angle ACB \\
 & \Rightarrow \angle ACB=\dfrac{\pi }{6}..........\left( 3 \right) \\
\end{align}$
Now let us consider the chord AC. We can see that the chord AC is subtending the angle $\angle AOC$ at the centre O and angle $\angle ABC$ at the point B.
Now let us consider the property that we have discussed above.
The angle subtended by any chord of the circle at the centre is double the angle subtended by the chord at any point on the circumference of the circle.
Using this property, we can write that
$\Rightarrow \angle AOC=2\angle ABC$
Substituting the value of $\angle AOC$ form equation (2), we get
$\begin{align}
  & \Rightarrow \dfrac{\pi }{4}=2\angle ABC \\
 & \Rightarrow \angle ABC=\dfrac{\pi }{8}..........\left( 4 \right) \\
\end{align}$
So, from equations (1), (3) and (4), we get that $\angle AOB=\dfrac{\pi }{3}$, $\angle ACB=\dfrac{\pi }{6}$ and $\angle ABC=\dfrac{\pi }{8}$.
Hence answer is \[\dfrac{\pi }{3},\dfrac{\pi }{6},\dfrac{\pi }{8}\].

Note: There is a possibility of one making a mistake by considering the property we have discussed in the solution in a wrong way as, the angle subtended by any chord of the circle at the circumference is double the angle subtended by the chord at the centre of the circle. We need to remember that angle subtended by the chord at the centre is always greater than the angle subtended by the chord at any point on the circle.