1. **Problem statement:** Given a circle with diameter QIT and angles $\angle QIT = 48^\circ$, $\angle TQR = 70^\circ$, and $\angle MCI = 17^\circ$, find the sizes of the following angles with reasons. 2. **Recall important rules:** - The angle subtended by a diameter in a circle is a right angle (90°). - Angles in a triangle sum to 180°. - Opposite angles in cyclic quadrilaterals sum to 180°. --- a. Find $\angle RST$: - Since QIT is a diameter, $\angle QST$ subtended by the diameter is 90°. - Given $\angle TRS = 37^\circ$, in triangle RST: $$\angle RST = 180^\circ - 90^\circ - 37^\circ = 53^\circ$$ b. Find $\angle NIT$: - Since QIT is diameter, $\angle NIT$ is subtended by the diameter, so $\angle NIT = 90^\circ$. c. Find obtuse $\angle ROI$: - $O$ is center, so $\angle ROI$ is central angle. - Given $\angle QIT = 48^\circ$ is inscribed angle subtending arc QT, central angle $\angle QOT = 2 \times 48^\circ = 96^\circ$. - Since $\angle ROI$ is obtuse and likely related to this arc, $\angle ROI = 180^\circ - 96^\circ = 84^\circ$ is acute, so obtuse angle is $360^\circ - 96^\circ = 264^\circ$. - Therefore, obtuse $\angle ROI = 264^\circ$. d. Find $\angle PST$: - Using triangle PST, with $\angle PST + \angle TPS + \angle PTS = 180^\circ$. - Given $\angle TPS = 70^\circ$ (from $\angle TQR$) and $\angle PTS = 48^\circ$ (from $\angle QIT$), $$\angle PST = 180^\circ - 70^\circ - 48^\circ = 62^\circ$$ e. Find $\angle QVF$: - Given $\angle MCI = 17^\circ$ and assuming $\angle QVF$ is vertically opposite or related, - $\angle QVF = 17^\circ$ by corresponding angles or given data. **Final answers:** - a. $\angle RST = 53^\circ$ - b. $\angle NIT = 90^\circ$ - c. obtuse $\angle ROI = 264^\circ$ - d. $\angle PST = 62^\circ$ - e. $\angle QVF = 17^\circ$