(a) PQ is a tangent to a circle RST at the point S. PRT is a straight line, < TPS = 34° and < TSQ = 65°. (i) Illustrate the information in a diagram; (ii) f...

Answer Details

(a) Tangent \(PQ\) touches the circle \(RST\) at \(S\), with \(PRT\) a straight secant, \(\angle TPS=34^\circ\) and \(\angle TSQ=65^\circ\).

(i) Diagram.

(ii)(a) Finding \(\angle RTS\). Since \(P\), \(S\) and \(Q\) lie on the tangent line, \(\angle TSQ=65^\circ\) is the exterior angle of triangle \(PST\) at \(S\). An exterior angle equals the sum of the two interior opposite angles, \(\angle TPS\) and \(\angle PTS\). Because \(R\) lies on \(PT\), the angle \(\angle PTS\) is the same as \(\angle RTS\):

\[ \angle TSQ=\angle TPS+\angle RTS \]

\[ 65^\circ=34^\circ+\angle RTS\ \Rightarrow\ \angle RTS=31^\circ. \]

(ii)(b) Finding \(\angle SRP\). By the alternate segment theorem, the tangent-chord angle \(\angle TSQ\) equals the inscribed angle in the alternate segment standing on chord \(ST\), which is \(\angle SRT\):

\[ \angle SRT=\angle TSQ=65^\circ. \]

Since \(PRT\) is a straight line, \(\angle SRP\) and \(\angle SRT\) are angles on a straight line:

\[ \angle SRP=180^\circ-\angle SRT=180^\circ-65^\circ=115^\circ. \]

(b) In the given diagram \(VZ=YZ\), \(\angle YXZ=20^\circ\) and \(\angle ZVY=52^\circ\); find \(\angle WYZ\).

Triangle \(VZY\) has \(VZ=YZ\), so it is isosceles and its base angles at \(V\) and \(Y\) are equal:

\[ \angle ZYV=\angle ZVY=52^\circ. \]

The inscribed angle \(\angle ZVY=52^\circ\) stands on chord \(ZY\), so the arc \(ZY\) it cuts off is

\[ \text{arc } ZY=2\times52^\circ=104^\circ. \]

\(X\) is an external point with the two secants \(XVY\) and \(XWZ\). The angle between two secants from an external point equals half the difference of the two intercepted arcs:

\[ \angle YXZ=\tfrac{1}{2}\big(\text{arc } YZ-\text{arc } VW\big) \]

\[ 20^\circ=\tfrac{1}{2}\big(104^\circ-\text{arc } VW\big)\ \Rightarrow\ \text{arc } VW=64^\circ. \]

Equal chords cut off equal arcs, so \(VZ=YZ\) gives \(\text{arc } VWZ=\text{arc } YZ=104^\circ\). Since \(W\) lies on the arc between \(V\) and \(Z\),

\[ \text{arc } WZ=\text{arc } VWZ-\text{arc } VW=104^\circ-64^\circ=40^\circ. \]

Finally, \(\angle WYZ\) is the inscribed angle at \(Y\) standing on chord \(WZ\), so it is half of arc \(WZ\):

\[ \angle WYZ=\tfrac{1}{2}\times40^\circ=20^\circ. \]