Question

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

• A. 40${}^o$
• B. 80${}^o$
• C. 50${}^o$
• D. 60${}^o$

Answer 1

Answer: (D)

60${}^o$

Angle subtended by chord BC at the centre will be double than that of angle

subtended by it at circumference.

therefore $\angle BOC=2\angle CAB=2(60{)}^{\circ }={120}^{\circ }$

and $\angle CAB=\angle CDB={60}^{\circ }$ {chord substend equal angle on circumference.}

$\angle COD={180}^{\circ }$.

Hence, $\angle COB+\angle BOD={180}^{\circ }$

Hence, $\angle BOD={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

In triangle BOD, $\angle BOD+\angle ODB+\angle DBO={180}^{\circ }$ (Angle sum property of a triangle)

Thus, ${60}^{\circ }+{60}^{\circ }+\angle ODB={180}^{\circ }$

Thus, $\angle ODB={60}^{\circ }$. Thus, $\angle ADB={60}^{\circ }$

So, option D is the answer.