Question

In the figure above ABCD is a cyclic quadrilateral and O is the centre of the circle B, O and D are collinear points If $\angle ABC={70}^{\circ }$ and $\angle OAD={50}^{\circ }$, then find $\angle OCB$.

• A. ${15}^{\circ }$
• B. ${20}^{\circ }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is C ${30}^{\circ }$

Given that:

$\angle ABC={70}^{\circ },\angle OAD={50}^{\circ }$

To find:

$\angle OCB=?$

Solution:

$\angle ADC+\angle ABC={180}^{\circ }$ (Sum of opposite angles of a cyclic quadrilateral is ${180}^{\circ }.$)

$\Rightarrow \angle ADC+{70}^{\circ }={180}^{\circ }$

$\Rightarrow \angle ADC={110}^{\circ }$

$\angle AOC=2\angle ABC$ (Angle at the centre is twice of angle subtended by the chord at the circumference.)

$\angle AOC=2\times {70}^{\circ }={140}^{\circ }$

Now, In quadrilateral $AOCD,$

$\angle AOC+\angle OAD+\angle ADC+\angle DCO={360}^{\circ }$

or, ${140}^{\circ }+{50}^{\circ }+{110}^{\circ }+\angle DCO={360}^{\circ }$

or, $\angle DCO={360}^{\circ }-{300}^{\circ }={60}^{\circ }$

Since, $B,O$ and $D$ are collinear

$\angle DCB={90}^{\circ }$

or, $\angle DCO+\angle OCB={90}^{\circ }$

or, $\angle OCB={90}^{\circ }-{60}^{\circ }$

or, $\angle OCB={30}^{\circ }$

Hence, C is the correct option.