(a) PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56°. Find (i) SQT (ii) PQT. (b) In the diagram, points B and ...

(a) \(PQRST\) is a circle, centre \(C\), with \(PCS\) a straight line (so \(PS\) is a diameter), \(RS \parallel QT\), \(|QR| = |RS|\), and \(\angle QTS = 56^\circ\).

Work with arcs (an inscribed angle equals half its intercepted arc).

\(\angle QTS = 56^\circ\) stands on chord \(QS\), intercepting arc \(QRS\):

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

Equal chords \(|QR| = |RS|\) cut off equal arcs, so

\[ \text{arc } QR = \text{arc } RS = \tfrac{112^\circ}{2} = 56^\circ \]

(i) \(\angle SQT\). Because \(RS \parallel QT\), the arcs intercepted between these parallel chords are equal, so \(\text{arc } ST = \text{arc } QR = 56^\circ\). Then \(\angle SQT\) stands on chord \(ST\):

\[ \angle SQT = \tfrac{1}{2}\,\text{arc } ST = \tfrac{1}{2}\times 56^\circ = 28^\circ \]

(ii) \(\angle PQT\). Since \(PS\) is a diameter, each semicircle is \(180^\circ\). Taking the semicircle \(S \to T \to P\):

\[ \text{arc } ST + \text{arc } TP = 180^\circ \;\Rightarrow\; 56^\circ + \text{arc } TP = 180^\circ \;\Rightarrow\; \text{arc } TP = 124^\circ \]

\(\angle PQT\) stands on chord \(PT\), intercepting arc \(TP\):

\[ \angle PQT = \tfrac{1}{2}\,\text{arc } TP = \tfrac{1}{2}\times 124^\circ = 62^\circ \]

(b) \(B\) and \(C\) are on the horizontal plane with \(|BC| = 30\) cm; \(A\) is vertically above \(B\) with \(|AB| = 26\) cm, and \(D\) is vertically above \(C\) with \(|DC| = 40\) cm.

(i) Angle of depression of \(B\) from \(D\). The horizontal from \(D\) is level with \(C\). \(B\) lies \(40\) cm below that level and \(30\) cm across:

\[ \tan\theta = \frac{DC}{BC} = \frac{40}{30} = 1.3333 \]

\[ \theta = \tan^{-1}(1.3333) = 53.13^\circ \approx 53^\circ \]

(ii) Angle of depression of \(A\) from \(D\). \(A\) is \(26\) cm above \(B\), so its top is \(40 - 26 = 14\) cm below the level of \(D\); the horizontal separation is still \(30\) cm:

\[ \tan\alpha = \frac{40 - 26}{30} = \frac{14}{30} = 0.4667 \]

\[ \alpha = \tan^{-1}(0.4667) = 25.02^\circ \approx 25^\circ \]

Answers: (a)(i) \(28^\circ\); (a)(ii) \(62^\circ\); (b)(i) \(53^\circ\); (b)(ii) \(25^\circ\).