Question

In the given figure O is the centre of the circle and AD = DC. If AB is the tangent to the circle and $\angle$ AOC = ${120}^{\circ }$. Find $\angle$ OCD.

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${45}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is A

${30}^{\circ }$

OA = OC (radius)
$\angle$ ADC = $\frac{1}{2}$ $\angle$AOC
$={60}^{\circ }$
In $\Delta$ADO and $\angle$CDO
AO = OC
OD = OD
AD = DC (given)
$\Delta$ ADO $\cong$ $\Delta$ COD
So $\angle$ ODC = ${30}^{\circ }$
$\angle$ OCD = ${30}^{\circ }$ (angles opposite to equal sides are equal)