Question: DELHI PUBLIC SCHOOL, GREATER NOIDA MATHEMATICS CLASS-IX CHAPTER 5 – I’m Up and Down, and Round and Round C.W Very Short Answer Type Questions 1. In the given figure, O is the centre of a circle such that $\angle OAB = 40^\circ$ and C is a point on the circle, then the value of $\angle ACB$ is A) $40^\circ$ B) $50^\circ$ C) $80^\circ$ D) $100^\circ$

1. **Problem:** Given a circle with center $O$, $\angle OAB = 40^\circ$, and $C$ is a point on the circle, find $\angle ACB$. 2. **Formula and Rule:** The angle subtended by a chord at the center is twice the angle subtended at any point on the remaining part of the circle. That is, if $\angle AOB$ is the central angle, then $\angle ACB = \frac{1}{2} \angle AOB$. 3. Since $O$ is the center, $OA = OB$ (radii), so triangle $OAB$ is isosceles. 4. Given $\angle OAB = 40^\circ$, then $\angle OBA = 40^\circ$ (base angles of isosceles triangle). 5. Sum of angles in triangle $OAB$ is $180^\circ$, so $$\angle AOB = 180^\circ - 40^\circ - 40^\circ = 100^\circ$$ 6. Using the rule, the angle subtended by chord $AB$ at point $C$ on the circle is half of $\angle AOB$: $$\angle ACB = \frac{1}{2} \times 100^\circ = 50^\circ$$ 7. **Answer:** $\boxed{50^\circ}$ (Option B)