a. Find the range of values of x which satisfy the following inequalities simultaneously: 5 - x > 1 and 9 + x \(\geq\) 8 In the diagram, O is the centre of ...

(a) Solving the inequalities simultaneously

First inequality:

\[ 5 - x > 1 \;\Rightarrow\; -x > 1 - 5 \;\Rightarrow\; -x > -4 \;\Rightarrow\; x < 4 \]

Second inequality:

\[ 9 + x \geq 8 \;\Rightarrow\; x \geq 8 - 9 \;\Rightarrow\; x \geq -1 \]

Combining both conditions:

\[ -1 \leq x < 4 \]

(b) Finding \(\angle QRS\)

From the diagram, \(O\) is the centre and \(PS\) passes through \(O\), so \(PS\) is a diameter. Also \(|PQ| = |QR|\) (equal chords, shown by the tick marks) and \(\angle PSR = 56^{\circ}\). The points \(P, Q, R\) lie on the arc on the same side of the diameter.

Step 1: Use equal chords. Equal chords subtend equal arcs, so \(\text{arc } PQ = \text{arc } QR\).

Step 2: Use the inscribed angle at S. \(\angle PSR = 56^{\circ}\) is the angle subtended at the circumference by the arc \(PQR\) (arc \(PR\) through \(Q\)). Hence

\[ \text{arc } PQR = 2 \times 56^{\circ} = 112^{\circ} \]

and since arc \(PQ = \) arc \(QR\), each equals \(56^{\circ}\).

Step 3: Find the remaining arc. Because \(PS\) is a diameter, the semicircle \(P\)-\(Q\)-\(R\)-\(S\) totals \(180^{\circ}\):

\[ \text{arc } RS = 180^{\circ} - \text{arc } PQ - \text{arc } QR = 180^{\circ} - 56^{\circ} - 56^{\circ} = 68^{\circ} \]

Step 4: Use the cyclic quadrilateral PQRS. Opposite angles of a cyclic quadrilateral are supplementary, so \(\angle QPS + \angle QRS = 180^{\circ}\).

\(\angle QPS\) subtends arc \(QRS = \text{arc } QR + \text{arc } RS = 56^{\circ} + 68^{\circ} = 124^{\circ}\), so

\[ \angle QPS = \tfrac{1}{2}\times 124^{\circ} = 62^{\circ} \]

Therefore

\[ \angle QRS = 180^{\circ} - 62^{\circ} = 118^{\circ} \]

\(\angle QRS = 118^{\circ}\).