Question

In the given figure, $BOD$ is the diameter of the circle with centre $O.$ $\angle COD={92}^{\circ }$ and $\angle ABD={65}^{\circ }.$ Then, $y$ equals

• A. ${65}^{\circ }$
• B. ${46}^{\circ }$
• C. ${44}^{\circ }$
• D. ${21}^{\circ }$

Answer 1

The correct option is D ${21}^{\circ }$

Given that:

$BOD$ is a diameter.

$\angle COD={92}^{\circ },\angle ABD={65}^{\circ }$

To find:

$y=?$

Solution:

In $△ABD$

$\angle ABD+\angle BAD+\angle ADB={180}^{\circ }$ (Sum of angles of a triangle is equal to ${180}^{\circ }.$)

or, ${65}^{\circ }+{90}^{\circ }+\angle ADB={180}^{\circ }$

or, $\angle ADB={180}^{\circ }-{155}^{\circ }={25}^{\circ }$

or, $\angle ADB=\angle ADE={25}^{\circ }$

Now,

$\angle CAD=\frac{\angle COD}{2}$ (Angle at the centre is twice of angle at the circumference by a chord.)

or, $\angle CAD=\angle EAD=\frac{{92}^{\circ }}{2}={46}^{\circ }$

Now,

$\angle CEO=\angle EAD+\angle ADE$ (Sum of remote interior angles is equal to exterior angle.)

or, $\angle CEO={46}^{\circ }+{25}^{\circ }={71}^{\circ }$

Again,

$\angle CEO+\angle ECO=\angle COD$ (Sum of remote interior angles is equal to exterior angle.)

or, ${71}^{\circ }+y={92}^{\circ }$

or, $y={21}^{\circ }$

Hence, D is the correct option.