In the diagram, PQRS is a circle. = . ?SPR = 26° and the interior angles of PQS are in the ratio 2:3 :3. Calculate: (i) PQR; (ii) RPQ; (iii) PRQ (b) The coo...

(a) Circle \(PQRS\) with \(|PS|=|PQ|\), \(\angle SPR=26^{\circ}\) and the interior angles of \(\triangle PQS\) in the ratio \(2:3:3\).

Step 1: Angles of triangle \(PQS\).

The three angles add up to \(180^{\circ}\). With \(2+3+3=8\) parts, one part is \(\dfrac{180^{\circ}}{8}=22.5^{\circ}\). Taking the angles at \(P:Q:S\) as \(2:3:3\):

\[\angle SPQ=2\times 22.5^{\circ}=45^{\circ}\]\[\angle PQS=3\times 22.5^{\circ}=67.5^{\circ}\]\[\angle PSQ=3\times 22.5^{\circ}=67.5^{\circ}\]

(The two equal angles \(67.5^{\circ}\) match the equal chords \(|PS|=|PQ|\) shown by the tick marks.)

Step 2: Arcs of the circle. Using "inscribed angle = half its arc":

• \(\angle SPR=26^{\circ}\) stands on arc \(SR\), so arc \(SR=52^{\circ}\).
• \(\angle PSQ=67.5^{\circ}\) stands on arc \(PQ\), so arc \(PQ=135^{\circ}\).
• \(\angle PQS=67.5^{\circ}\) stands on arc \(PS\), so arc \(PS=135^{\circ}\).
• The four arcs total \(360^{\circ}\): arc \(RQ=360^{\circ}-(135^{\circ}+52^{\circ}+135^{\circ})=38^{\circ}\).

(i) \(\angle PQR\).

\[\angle PQR=\angle PQS+\angle SQR\]

\(\angle SQR\) stands on arc \(SR=52^{\circ}\), so \(\angle SQR=26^{\circ}\). Hence:

\[\angle PQR=67.5^{\circ}+26^{\circ}=93.5^{\circ}\]

(ii) \(\angle RPQ\).

\[\angle RPQ=\angle SPQ-\angle SPR=45^{\circ}-26^{\circ}=19^{\circ}\]

(iii) \(\angle PRQ\). \(\angle PRQ\) and \(\angle PSQ\) both stand on the same chord \(PQ\) (angles in the same segment), so they are equal:

\[\angle PRQ=\angle PSQ=67.5^{\circ}\]

Check: in cyclic quadrilateral \(PQRS\), \(\angle PQR+\angle PSR=93.5^{\circ}+(67.5^{\circ}+19^{\circ})=93.5^{\circ}+86.5^{\circ}=180^{\circ}\). Correct.

(b) Points \(P(7,3)\) and \(Q(5,x)\) with \(|PQ|=\sqrt{29}\) units.

Distance formula:

\[|PQ|^2=(7-5)^2+(3-x)^2\]\[29=2^2+(3-x)^2\]\[29=4+(3-x)^2\]\[(3-x)^2=25\]\[3-x=\pm 5\]

So \(3-x=5\Rightarrow x=-2\), or \(3-x=-5\Rightarrow x=8\).

\[x=8 \quad\text{or}\quad x=-2\]