Question

In the following figure, O is the center of circle. Find the value of $\angle ADB$.

• A. 70°

Answer 1

Answer: (A)

70°

Answer 2

To find the value of $\angle ADB$, we will use properties of circles and triangles.

1. Interpret the given angles: The angles marked 40° and 30° in the figure are $\angle OAC$ and $\angle OBC$ respectively. This is inferred from the placement of the angle arcs and the lines involved.

2. Use properties of isosceles triangles formed by radii:

• In $\triangle OAC$: OA and OC are radii of the same circle, so OA = OC. Therefore, $\triangle OAC$ is an isosceles triangle. The angles opposite to equal sides are equal, so $\angle OCA = \angle OAC$. Given $\angle OAC = 40^\circ$, thus $\angle OCA = 40^\circ$.

• In $\triangle OBC$: OB and OC are radii of the same circle, so OB = OC. Therefore, $\triangle OBC$ is an isosceles triangle. The angles opposite to equal sides are equal, so $\angle OCB = \angle OBC$. Given $\angle OBC = 30^\circ$, thus $\angle OCB = 30^\circ$.

3. Calculate $\angle ACB$: The angle $\angle ACB$ is the sum of $\angle OCA$ and $\angle OCB$. $\angle ACB = \angle OCA + \angle OCB$ $\angle ACB = 40^\circ + 30^\circ$ $\angle ACB = 70^\circ$

4. Use the property of angles in the same segment: Angles subtended by the same arc at any point on the remaining part of the circle are equal. In the given figure, arc AB subtends $\angle ACB$ at point C and $\angle ADB$ at point D on the circumference. Therefore, $\angle ADB = \angle ACB$.

5. Final Value: Since $\angle ACB = 70^\circ$, we have: $\angle ADB = 70^\circ$