Question

In the figure, \[A\], \[B\], and \[C\] are three points on the circle with centre \[O\] such that \[\angle BOC = 30^\circ \] and \[\angle AOB = 60^\circ \]. If \[D\] is another point on the circle other than the arc \[ABC\], find \[\angle ADC\].

Answer 1

Hint: Here, we will use the theorem for angle subtended by the centre to find the measure of \[\angle ADC\]. The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc on any other point of the same circle.

Complete step-by-step answer:
First, we will find the measure of angle \[AOC\].
From the figure, we can observe that the measure of angle \[AOC\] is the sum of the measures of angle \[BOC\] and angle \[AOB\].
Therefore, we get
\[\angle AOC = \angle AOB + \angle BOC\]
Substituting \[\angle AOB = 60^\circ \] and \[\angle BOC = 30^\circ \] in the equation, we get
\[ \Rightarrow \angle AOC = 60^\circ + 30^\circ \]
Adding the terms in the expression, we get
\[ \Rightarrow \angle AOC = 90^\circ \]
Now, we will use the theorems of a circle to find the measure of angle \[ADC\].
We know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the same arc on any other point of the circle.
The angle subtended by the arc \[ABC\] at the centre of the circle is angle \[AOC\].
The angle subtended by the arc \[ABC\] at the point \[D\] on the circle is angle \[ADC\].
Therefore, using the theorem, we get
\[\angle AOC = 2\angle ADC\]
Substituting \[\angle AOC = 90^\circ \] in the equation, we get
\[ \Rightarrow 90^\circ = 2\angle ADC\]
Dividing both sides of the equation by 2, we get
\[ \Rightarrow \dfrac{{90^\circ }}{2} = \dfrac{{2\angle ADC}}{2}\]
Therefore, we get
\[ \Rightarrow \angle ADC = 45^\circ \]
\[\therefore \] We get the measure of angle \[ADC\] as \[45^\circ \].

Note: A common mistake is to use the theorem to write \[2\angle AOC = \angle ADC\]. This leads to the incorrect answer \[\angle ADC = 180^\circ \]. It is clear from the figure that \[\angle ADC\] cannot measure \[180^\circ \].