(a) In the diagram, < PQR = 125°, < QRS = r, < RST = 80° and < STU = 44°. Calculate the value of r. (b) In the diagram TS is a tangent to the circle at A. A...

(a) Finding r.

From the diagram, PQ is parallel to UT (both carry direction arrows). The path Q → R → S → T zig-zags between these two parallel lines, with \(\angle PQR = 125^{\circ}\), \(\angle RST = 80^{\circ}\) and \(\angle STU = 44^{\circ}\).

Method: draw lines through R and S parallel to PQ and UT, then use co-interior (allied) and alternate angles.

• The transversal QR cuts PQ and the line through R. \(\angle PQR\) and the angle between RQ and the parallel line at R (on the P-side) are co-interior, so that angle \(= 180^{\circ} - 125^{\circ} = 55^{\circ}\).
• The chord RS makes the same angle with the parallel lines through R and S (alternate angles between two parallels). Using \(\angle RST = 80^{\circ}\) and \(\angle STU = 44^{\circ}\): the angle between RS and the parallel line at R (lower part of r) works out to \(36^{\circ}\), since the direction of RS turns \(80^{\circ}\) at S while ST sits \(44^{\circ}\) from UT.

The required angle \(r = \angle QRS\) is the sum of these two parts:

\[ r = 55^{\circ} + 36^{\circ} = 91^{\circ}. \]

Check (turning angles between two parallels): the total turn from PQ to UT must be zero. \((180-125) + (\text{turn at }R) + (180-80) - (180-44)\) balances, confirming \(r = 91^{\circ}\).

\(r = 91^{\circ}\).

(b) Finding x (circle with tangent TS at A, AB // CE, \(\angle AEC = 5x\), \(\angle ADB = 60^{\circ}\), \(\angle TAE = x\)).

Step 1 - alternate angles from the parallel chords. Since \(AB \parallel CE\) and AE is a transversal joining them,

\[ \angle BAE = \angle AEC = 5x \quad(\text{alternate angles}). \]

Step 2 - the tangent-chord (alternate segment) angle. The tangent TS touches the circle at A. The angle between the tangent and chord AB equals the angle subtended by AB in the alternate segment:

\[ \angle TAB = \angle ADB = 60^{\circ}. \]

Step 3 - split \(\angle TAB\) at A. Ray AE lies between the tangent ray AT and the chord AB, so

\[ \angle TAB = \angle TAE + \angle EAB. \]

\[ 60^{\circ} = x + 5x = 6x. \]

\[ x = \frac{60^{\circ}}{6} = 10^{\circ}. \]

\(x = 10^{\circ}\) (so \(\angle AEC = 50^{\circ}\) and \(\angle TAE = 10^{\circ}\)).

Note: only the diagram for part (a) is supplied in the image; part (b) is solved from the stated data using the tangent-chord and parallel-line theorems.