1. **Problem statement:** Given a circle with center O, OA = 5 cm (radius), AB = 12 cm, and AB and BC are tangents to the circle at points A and C respectively. (i) Find the size of angle OAB. (ii) Find the length of OB. (iii) Given OA is a radius and PB is tangent at C, find angle COB. 2. **Step-by-step solution:** **(i) Calculate angle OAB:** - Since AB is tangent at A, OA is radius, so OA \perp AB. - Triangle OAB is right angled at A. - Using Pythagoras theorem in triangle OAB: $$OB^2 = OA^2 + AB^2 = 5^2 + 12^2 = 25 + 144 = 169$$ $$OB = \sqrt{169} = 13 \text{ cm}$$ - To find angle OAB, use trigonometry: $$\tan(\angle OAB) = \frac{OA}{AB} = \frac{5}{12}$$ $$\angle OAB = \tan^{-1}\left(\frac{5}{12}\right) \approx 22.62^\circ$$ **(ii) Length of OB:** - From above, $$OB = 13 \text{ cm}$$ **(iii) Calculate angle COB:** - Since PB is tangent at C, and OC is radius, angle between radius and tangent is 90°. - Triangle OCB is right angled at C. - OB = 13 cm (from above), OC = OA = 5 cm (radius). - Use cosine rule or trigonometry: $$\cos(\angle COB) = \frac{OC}{OB} = \frac{5}{13}$$ $$\angle COB = \cos^{-1}\left(\frac{5}{13}\right) \approx 67.38^\circ$$ --- 4. **Problem statement:** Given circle with center O, angle \(\angle POQ = 84^\circ\), points P, Q, R on circumference. (a) Find \(\angle PRQ\). (b) Find \(\angle OQR\). **Step-by-step solution:** (a) \(\angle PRQ\) is an angle at the circumference subtended by arc PQ. - Angle at center \(\angle POQ = 84^\circ\). - Angle at circumference subtending same arc is half the central angle: $$\angle PRQ = \frac{1}{2} \times 84^\circ = 42^\circ$$ (b) To find \(\angle OQR\): - Triangle OQR is isosceles with OP = OQ = radius. - \(\angle POQ = 84^\circ\) given. - Sum of angles in triangle OQR is 180°. - \(\angle OQR = \angle ORQ\) (since OQ = OR). - Let \(\angle OQR = x\). - Then: $$84^\circ + x + x = 180^\circ$$ $$2x = 96^\circ$$ $$x = 48^\circ$$ **Final answers:** (i) \(\angle OAB \approx 22.62^\circ\) (ii) \(OB = 13 \text{ cm}\) (iii) \(\angle COB \approx 67.38^\circ\) (4a) \(\angle PRQ = 42^\circ\) (4b) \(\angle OQR = 48^\circ\)