out of the two concentric circles, the radius of the outer circle is 5 cm and the chord ac of length 8 cm is a tangent to the inner circle. find the radius of the inner circle.
let c1 and c2 are two concentric circles with center o and radius of outer circle is 5 cm.given : ac is a chord with length 8 cm that is tangent to inner circle .to find : radius of inner circle i.e. odac is a tangent for c1 at d so,od ⏊ ac [tangent at a point on the circle is perpendicular to the radius through point of contact ]so, oad is a right-angled triangle at d .also it implies that od is perpendicular to the chord ac in c2so we have,ad = dc [perpendicular from the center to the chord bisects the chord]ac = ad + dc8 = ad + adad = 4 cmin △oad by pythagoras theorem(oa)2 = (od)2 + (ad)2(5)2 = (od)2+ (4)225 = (od)2 + 16(od)2= 9od = 3 cm
consider the following figure:two circles with centres q and r touch each other externally at x. another circle, with centre p, touches the other two circles internally at s and t. if pq = 8 cm, qr = 11 cm and pr = 9 cm, find the radius of circle r. you can find the radius in the following way. 
write the equation of the line that is perpendicular to the given tangent line. this perpendicular goes through the stated center, and intersects the tangent; the perpendicular's slope is the negative reciprocal of the stated tangent line.
find the intersection (point t) of the tangent and radius line--then use the distance formula to find the radius, which is a segment from the stated center to the point t. sin(α β) = sinαcosβ cosαsinβ 
sin(α-β) = sinαcosβ - cosαsinβ


cos(α β) = cosαcosβ - sinαsinβ
cos(α-β) = cosαcosβ sinαsinβ


tan(α β) = tanα tanβ / 1-tanα tanβ
tan(α-β) = tanα-tanβ / 1 tanα tanβ


sin2θ = 2sinθcosθ = 2tanθ /1 tanθtanθ


cos2θ = cos平方θ−sin平方θ = 2cos平方θ−1 = 1−2sin平方θ


△abc=海龍公式 √s(s-a)(s-b)(s-c) 
[s=a b c/2 ，就是周長/2 ]



△abc=1/2 ａ.b*sinc = axbxc / 4r = rxs 

 [r=外接圓半徑，r=內接圓半徑]




應該是這樣吧!希望對你有幫助。 圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872668990.jpg
 
首先請先 複習下列公式
 
設扇形的中心角θ弳，半徑為r，弧長為s，面積為a，則
s=rθ；a= r2θ= rs
sol: 設: ∠aob=θ 弳 (弧度) od=x
(x+2.5) θ =6 ……………………………..(1)
求黃色部分面積= (x+5)2 θ - x2θ
= θ ( x2+10x+25-x2 )
= θ (10x+25)
=5θ( x + 2.5)=30



圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669011.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669012.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669033.jpg
 



 2010-06-04 23:25:06 補充：
y= |cosx|

恕不再多畫 將以文字說明

1. 當cosx >0 時 y=cosx
2. 當cosx <0 時 y = -cosx

 2010-06-04 23:26:21 補充：
注意

y=3sin2(x-π)+2

之 a , b 步驟為同一圖形

 2010-06-05 13:06:49 補充：
奇怪 1/2 通通不見了

更正

s=rθ；a=0.5 r2θ=0.5 rs

求黃色部分面積=0.5(x+5)2 θ - 0.5x2 θ
=0.5θ ( x2+10x+25-x2 )
=θ (10x+25)
=5θ( x + 2.5)=30

p.s. x2 即為 x平方

 2010-08-28 11:35:07 補充：
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rid of answer: `1.8706 xx 10^(11) " erg", 4477 cal` The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 solution: equation of parabola `y^(2)=8x` . . .(1)   
 comparing with `y^(2)=4ax`   
 4a=8   
 `rarr" "a=2`   
 co-ordinates of ends of focal chord of parabola (1)   
 `=(at^(2),2at)and((a)/(t^(2)),(-2a)/(t))`   
 `=(2t^(2),4t)and((2)/(t^(2)),(-4)/(t))`   
 the mid-point of this chord is (2,0).   
 `:." "(2t^(2)+(2)/(t^(2)))/(2)=2and(4t-(4)/(t))/(2)=0`   
 `rarr" "t^(2)+(1)/(t^(2))=andt-(1)/(t)=0`   
 `rarr" "t=1`   
 therefore, the co-ordinates of the ends of latus rectum=(2,4) and (2,-4)   
 `:.` equation of latus rectum   
 `y-4=(-4-4)/(2-2)(x-2)`   
 `rarr" "x-2=0`. you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) [readmore]typed answerh2 + i2 ⇌ 2hi      kc = 4.6i            0.5   0.5     0.5c          -x      -x       +2xe        0.5-x 0.5-x   0.5+2xkc=hi2hii2kc=0.5+2x20.5-x0.5-xkc=0.5+2x20.5-x24.6=0.5+2x0.5-xx=0.138hi=0.5+2x=0.776 m a=c and b=d
b=d and `anec`
`ab=1 " and "cdne1`
`cd=1" and "ab=ne1`
answer: a
solution: using `a.m.gtg.m,` we have   
 `a+(1)/(b)gt2sqrt((a)/(b))b+(1)/( c )gt2sqrt((b)/( c )),c+(1)/(d)gt2sqrt(( c )/(d))" and "d+(1)/(a)gt2sqrt((d)/(a))`   
 `implies(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))gt16`   
 but,`(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))=4xx1xx4xx1=16`   
 `:." "a=(1)/(b),b=(1)/( c ),c=(1)/(d)" and "d=(1)/(a)`   
 `implies" "a=(1)/(b)=2,b=(1)/( c )=(1)/(2),c=(1)/(d)=2" and "d=(1)/(a)=(1)/(2)`   
 `implies" "a=2,b=(1)/(2),c=2" and "d=(1)/(2)`   
 `implies" "a=c" and "b=d`