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!
F.5 Maths Circle(5)
2014-12-27 11:25 pm
回答 (2)
2014-12-28 3:05 am
✔ 最佳答案
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
參考: ME
2014-12-28 12:43 am
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)
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)
收錄日期: 2021-04-15 17:40:30
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