1. **Problem Statement:** Given a circle with center O and points A, B, C on the circumference, angle AOB at the center is $t$. Angles on the circumference are $\angle ABC = 38^\circ$ and $\angle BCA = 24^\circ$. Find the value of $t$. 2. **Relevant Theorem:** The central angle $t$ subtending an arc is twice any inscribed angle subtending the same arc. That is, if $t$ is the central angle for arc AC, then $t = 2 \times \angle ABC$ or $t = 2 \times \angle BCA$ depending on the arc. 3. **Step 1: Find $\angle BAC$** Since $\triangle ABC$ is a triangle, sum of angles is $180^\circ$: $$\angle ABC + \angle BCA + \angle BAC = 180^\circ$$ $$38^\circ + 24^\circ + \angle BAC = 180^\circ$$ $$\angle BAC = 180^\circ - 62^\circ = 118^\circ$$ 4. **Step 2: Identify the arc subtended by $t$** Angle $t$ is the central angle $\angle AOB$ subtending arc AB. 5. **Step 3: Use the Inscribed Angle Theorem** The inscribed angle $\angle ACB$ subtends the same arc AB as the central angle $t$. 6. **Step 4: Calculate $t$** By the theorem: $$t = 2 \times \angle ACB$$ We know $\angle BCA = 24^\circ$, so: $$t = 2 \times 24^\circ = 48^\circ$$ **Final answer:** $$t = 48^\circ$$