AC是圆O的直径,PA,PB是圆O的切线。A,B为切点,AB=6,PA=5.求圆O的半径
求:(1)圆O的半径(2)sin∠BAC的值(答案:第一题是4分之15,第二题是5分之3,,,我要过程!!!)...
求:(1)圆O的半径
(2)sin∠BAC的值
(答案:第一题是4分之15,第二题是5分之3,,,我要过程!!!) 展开
(2)sin∠BAC的值
(答案:第一题是4分之15,第二题是5分之3,,,我要过程!!!) 展开
1个回答
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AB = 6, PA =5
let OB =OA = r and ∠AOB = 2x
=> ∠AOP = ∠POB = x
consider △AOB
sin (90°-x)/r = sin2x/AB
cosx/r = sin2x/6
r = 6cosx/sin2x
= 3/sinx (1)
consider △OAP
sinx = AP/OP
= AP/√(AP^2+OA^2)
= 5/√(25+r^2) (2)
sub (2) into (1)
r(5/√(25+r^2) ) = 3
5r = 3√(25+r^2)
25r^2 = 9(25 +r^2)
r^2 = 9(25)/16
r = 15/4
(2)
BC^2 + AB^2 = OC^2
BC^2+ 36 = 225/4
BC = 9/2
sin∠BAC = BC/AC
= BC/2r
= 2BC/15
= 3/5
let OB =OA = r and ∠AOB = 2x
=> ∠AOP = ∠POB = x
consider △AOB
sin (90°-x)/r = sin2x/AB
cosx/r = sin2x/6
r = 6cosx/sin2x
= 3/sinx (1)
consider △OAP
sinx = AP/OP
= AP/√(AP^2+OA^2)
= 5/√(25+r^2) (2)
sub (2) into (1)
r(5/√(25+r^2) ) = 3
5r = 3√(25+r^2)
25r^2 = 9(25 +r^2)
r^2 = 9(25)/16
r = 15/4
(2)
BC^2 + AB^2 = OC^2
BC^2+ 36 = 225/4
BC = 9/2
sin∠BAC = BC/AC
= BC/2r
= 2BC/15
= 3/5
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