A circular flower bed has centre \(O\) and radius \(20\) m. The sponsor's tent, at point \(T\) outside the bed, is joined to the edge by ropes \(TA\) and \(...

(a) \(OA\) is a radius and \(AT\) is tangent to the bed at \(A\), so angle \(OAT = 90^{\circ}\), a tangent always meeting its radius at right angles. [1 mark]

(b) Triangle \(OAT\) is right-angled at \(A\), so by Pythagoras, \(OT = \sqrt{OA^{2}+AT^{2}} = \sqrt{20^{2}+24^{2}} = \sqrt{976} = 31.2\) m, correct to 3 significant figures. Using the same triangle, \(\tan(\text{angle }ATO) = \dfrac{OA}{AT} = \dfrac{20}{24}\), so angle \(ATO = \tan^{-1}(0.8\overline{3}) = 39.8^{\circ}\), correct to 3 significant figures. [3 marks]

(c) The figure is symmetrical about line \(OT\), so angle \(ATB\) is twice angle \(ATO\): angle \(ATB = 2 \times 39.8 = 79.6^{\circ}\). [2 marks]

(d) Quadrilateral \(OATB\) has angles summing to \(360^{\circ}\): two right angles at \(A\) and \(B\), plus \(79.6^{\circ}\) at \(T\), leaves angle \(AOB = 360-90-90-79.6 = 100.4^{\circ}\). [1 mark]

(e) \(A\), \(B\) and \(C\) all lie on the flower bed's circle, and \(C\) is on the major arc \(AB\), so the angle at the centre, angle \(AOB\), is twice the angle at the circumference, angle \(ACB\): angle \(ACB = 100.4 \div 2 = 50.2^{\circ}\), correct to 3 significant figures. [1 mark]

(f) No. \(O\) is the centre of the circle on which \(A\), \(B\) and \(C\) lie, and the centre of a circle is never a point on the circle itself, since every point on the circle is a full radius away from the centre, not zero. [1 mark]

Part (f) checks a conceptual point rather than a calculation: a "centre" and a "point on the circumference" are different roles, so \(O\) cannot join \(A\), \(B\), \(C\) on their circle even though it governs the angles between them.