Question

In the given figure, $AOB$ and $COD$ are two diameters of a circle with center O. If $\angle OAC = {60^0}$ then $\angle ABD$ is

\[
  A.{\text{ }}{40^0} \\
  B.{\text{ }}{80^0} \\
  C.{\text{ }}{50^0} \\
  D.{\text{ }}{60^0} \\
 \]

Answer 1

Hint: In order to solve the given problem and to find the given angle we will use the properties of angle subtended by chord in a circle. We know that angles subtended by chords on two different points in a circle are the same and angle subtended by the chord on the centre is double the angle subtended by the chord on the other point of circle. We will use these properties to solve the problem.

Complete step by step answer:
Given that:
$AOB$ and $COD$ are two diameters of a circle with center O.
And also we have:
$\angle OAC = {60^0}$
We have to find the value of $\angle ABD$
We will solve the problem according to the given figure.
We will use the properties of angles subtended by chords to find different angles of triangles in the figure.
Let us consider chord BC from the figure.
As we know that the angle subtended by any chord on the center of the circle is double the angle subtended by the same chord on any other point on the circumference of the circle.
So we have:
$\angle BOC = 2\angle BAC = 2 \times {60^0} = {120^0}$
By the linear pair property we have:
$
   \Rightarrow \angle BOC + \angle COA = {180^0} \\
   \Rightarrow {120^0} + \angle COA = {180^0} \\
   \Rightarrow \angle COA = {180^0} - {120^0} \\
   \Rightarrow \angle COA = {60^0} \\
 $
As angle $\angle DOB$ and angle $\angle COA$ are vertically opposite angles so they are equal. As the vertically opposite angles are always equal. So, we have:
$\angle DOB = \angle COA = {60^0}$
Now as we know that the angle subtended by any particular chord at two different places on the same side of the chord are always equal. So, we have:
$\angle CAB = \angle CDB = {60^0}$
Where angle $\angle CAB$ and angle $\angle CDB$ are two angles subtended by chord BC on the same side of the chord.
Now in the triangle $\Delta DOB$ we know the value of angle $\angle DOB$ and angle $\angle ODB$ . So let us now find the value of angle $\angle OBD$ .
As we know that the sum of all the three angles of a triangle is ${180^0}$ .
So, we have:
\[
   \Rightarrow \angle DOB + \angle ODB + \angle OBD = {180^0} \\
   \Rightarrow {60^0} + {60^0} + \angle OBD = {180^0} \\
   \Rightarrow \angle OBD = {180^0} - \left( {{{60}^0} + {{60}^0}} \right) \\
   \Rightarrow \angle OBD = {180^0} - \left( {{{120}^0}} \right) \\
   \Rightarrow \angle OBD = {60^0} \\
   \Rightarrow \angle ABD = {60^0} \\
 \]
Hence, in the given figure we have the value of angle \[\angle ABD = {60^0}\]
So, option D is the correct answer.

Note: In order to solve such problems, students must use the figure and must remember basic theorems of relating the chord and the angle subtended by it on the circle. This problem can also be solved by finding different angles of the first triangle, proving the triangles similar or the chords parallel and further finding the angle but the method used above is the simplest. Students must remember that the angles subtended by chords only on one side will be equal and not on the other side as for the other side there are other relations like linear pairs.