Question

In the given figure, O is the centre of the circle. Find \[\angle CBD\] .

Answer 1

Hint: At first find the value of $\angle APC$ by using the property that the angle subtended by the arc at the centre is twice the angle subtended at the circumference. Then use the property of cyclic quadrilateral where the sum of OP positive angles is $180{}^\circ $to find $\angle ABC$ . They use the fact that the angle in a straight line is $180{}^\circ $ . We can find the value of $\angle CBD$ .

Complete step-by-step answer:

In the question a figure is given \[\angle CBD\] if given angle AOC is 100.

So, given angle $\angle AOC$ is $100{}^\circ $. We can say that the angle at the centre of the circle is $100{}^\circ $. We know a property of a circle is that the angle subtended by an arc at the centre is twice the angle subtended at the circumference of the circle.

The angle subtended by an arc at the centre of the circle $100{}^\circ $. So, the angle subtended at the circumference will be $\dfrac{100{}^\circ }{2}$ or $50{}^\circ $ . So, here the angle at the circumference is $\angle APC$ .Now, we can say that $\angle APC$ is $50{}^\circ $.

If the endpoints of a quadrilateral lie in the circumference of a circle then we can say that the quadrilateral shows properties of a cyclic quadrilateral. If a quadrilateral is a cyclic quadrilateral then its sum of pairs of opposite angles add up to $180{}^\circ $ .

So, $\angle APC+\angle ABC=180{}^\circ $ .

We know that $\angle APC=50{}^\circ $ .

So, we can say, $\angle ABC=180{}^\circ -50{}^\circ =130{}^\circ $.

So, as we know that

ABD is a straight line and angle subtended in a straight line is $180{}^\circ $ . So, we can say that –

$\angle CBA+\angle CBD=180{}^\circ $ .

We found out that $\angle CBA$ is \[130{}^\circ \] .

Then, $\angle CBD=180{}^\circ -130{}^\circ $ or $50{}^\circ $.

Here angle $\angle CBD$ is $50{}^\circ $.


Note: After finding $\angle APC$ we can directly say the value of $\angle CBD$ by the property which states, the exterior angle of the cyclic quadrilateral is equal to the interior opposite angle. So, we can directly say that $\angle CBD$ is $50{}^\circ $.