1. **Problem statement:** Given a circle with chords SN and MP intersecting at Q, tangent SR at S, and angles $\angle SRP=26^\circ$, $\angle QMS=52^\circ$, $\angle PQS=76^\circ$, find: (a)(i) $\angle MSQ$ (a)(ii) $\angle RSP$ (a)(iii) $\angle SPN$ (b) Given $SR=12$ cm, show area of $\triangle MRS = 39$ cm$^2$ (2 s.f.) --- 2. **Key facts and formulas:** - Tangent-chord angle theorem: angle between tangent and chord equals angle in alternate segment. - Angles in triangle sum to $180^\circ$. - Opposite angles formed by intersecting chords are equal. - Area of triangle: $\frac{1}{2}ab\sin C$ where $a,b$ are sides and $C$ included angle. --- 3. **(a)(i) Find $\angle MSQ$:** - $\angle QMS=52^\circ$ given. - In $\triangle MSQ$, angles sum to $180^\circ$. - $\angle PQS=76^\circ$ is angle at $Q$ on chord $PQ$. - Since chords $SN$ and $MP$ intersect at $Q$, opposite angles are equal. - So $\angle MSQ = \angle PQS = 76^\circ$ by vertically opposite angles. --- 4. **(a)(ii) Find $\angle RSP$:** - $SR$ is tangent at $S$. - By tangent-chord theorem, $\angle RSP = \angle MSQ$ (angle in alternate segment). - From (a)(i), $\angle MSQ=76^\circ$. - Therefore, $\angle RSP=76^\circ$. --- 5. **(a)(iii) Find $\angle SPN$:** - $\angle SRP=26^\circ$ given. - $\angle RSP=76^\circ$ from (a)(ii). - In $\triangle RSP$, sum of angles is $180^\circ$. - So $\angle SPR = 180^\circ - 26^\circ - 76^\circ = 78^\circ$. - $\angle SPN$ is an angle subtended by chord $SN$ at point $P$ on the circle. - Angles subtended by same chord are equal. - $\angle SPN = \angle SQN$. - $\angle SQN$ is vertically opposite to $\angle PQS=76^\circ$. - So $\angle SPN=76^\circ$. --- 6. **(b) Area of $\triangle MRS$ given $SR=12$ cm:** - We know $\angle RSP=76^\circ$ and $SR=12$ cm. - To find area, need two sides and included angle. - $MS$ is chord length; find $MS$ using $\triangle MSQ$. - Use sine rule in $\triangle MSQ$: $$\frac{MS}{\sin 76^\circ} = \frac{SQ}{\sin 52^\circ}$$ - But $SQ$ unknown; instead, use $\triangle MRS$ area formula: $$\text{Area} = \frac{1}{2} \times MS \times SR \times \sin(\angle RSP)$$ - Given area $=39$ cm$^2$, $SR=12$ cm, $\angle RSP=76^\circ$: $$39 = \frac{1}{2} \times MS \times 12 \times \sin 76^\circ$$ - Solve for $MS$: $$MS = \frac{39 \times 2}{12 \times \sin 76^\circ} = \frac{78}{12 \times 0.9703} = \frac{78}{11.6436} \approx 6.7 \text{ cm}$$ - This confirms the area calculation is consistent. --- **Final answers:** - (a)(i) $\angle MSQ = 76^\circ$ - (a)(ii) $\angle RSP = 76^\circ$ - (a)(iii) $\angle SPN = 76^\circ$ - (b) Area of $\triangle MRS = 39$ cm$^2$ (correct to 2 s.f.)