標題:
F.5 Maths Circle(5)
發問:
Picture: http://postimg.org/image/850gr2tez/In the figure, PABQ is a common tangent to two circles with centres O1 and O2, where A and B are their respective points of contact. If the radii od two circles are 10cm and 6cm, and O1O2=18cm, find the length of AB.My calculation:... 顯示更多 Picture: http://postimg.org/image/850gr2tez/ In the figure, PABQ is a common tangent to two circles with centres O1 and O2, where A and B are their respective points of contact. If the radii od two circles are 10cm and 6cm, and O1O2=18cm, find the length of AB. My calculation: http://postimg.org/image/5cbuksigb/ I cannot continue to calculate. I want to ask is there any methods to find angle O1O2B? If no, how should I solve this question? Please help, thank you!
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最佳解答:
Add a line connect O2 to Radius of O1 and perpendicular to O1 and form a right-angled triangle. The height of the triangle: Radius of O1 - Radius of O2 =10-6 =4cm The length of the base of triangle=AB & O1O2=18cm. Therefore, by Pyth. theorem, 4^2 + AB^2 = 18^2 AB^2 = 308 AB=17.5cm (corr. to 3 sig. fig.) 2014-12-27 19:07:47 補充： perpendicular to the radius of O1* 2014-12-27 19:12:56 補充： To find ∠O1O2B: Let the point that radius of O1 perpendicular to the radius of O2 be C. sin ∠O1O2C = 4/18 ∠O1O2C = 12.8 (corr. to 3 sig. fig.) Therefore, ∠O1O2B= ∠O1O2C + 90 ∠O1O2B=102.8
其他解答:
你或且可以將AB平行地拉至到O2到 2014-12-27 16:57:03 補充： Method 2 you can extend O1O2 to C where C is on PQ s.t. O1O2C is a straight line then, using similiar triangle's property, O1C/O2C=O1A/O2B (18+O2C)/O2C=10/6=5/3 O2C=27 cos ∠BO2C=6/27 ∠BO2C=arccos (6/27) ∠O1O2B=180°- arccos (6/27)
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