trigonometric ratio~f.2今日要!緊急

1 Add a centre O.
Then angle APO= angle PAO
angle angle BPO= angle PBO
Also angle APO + angle PAO + angle BPO + angle PBO = 180
So angle APO + angle BPO = 90
That is angle P = 90
PN= PA sin theta
PN/PB=cos theta => PB=PN/cos theta = a tan theta
2 We have
tan alpha = ON/(2PQ+RN)
tan beta = ON/(PQ+RN)
tan x = ON/RN
Rearranging
1/tan alpha = (2PQ+RN)/ON
1/tan beta = (PQ+RN)/ON
1/tan x = RN/ON
So 1/tan alpha = 2/tan beta-1/tanx
1/tan alpha + 1/tanx = 2/tan beta


2009-04-22 18:56:29 補充：
q1: NB=PB sin theta = a tan theta sin theta

Q1. Angle APB = 90 degree is a property of semicircle in geometry.
By similar triangle angle BPN = angle PAB = x ( instead of using theta).
PN = a sin x
so NB = PN tan x = a sin x tan x.
Q2.
By sine rule:
OP/sin x = OR/sin a
OP/OR = sin x/sin a..................(1)
PQ/sin ( b - a) = OP/sin b............(2)
QR/sin ( x - b) = OR/sin b............(3)
(2)/(3) we get
PQ sin ( x - b)/QR sin ( b -a) = OP/OR
since PQ = QR therefore,
sin ( x - b)/sin ( b - a) = OP/OR = sin x/sin a.

圖片參考：http://farm4.static.flickr.com/3518/3464572945_7d77f9ca32.jpg?v=0

2)Let O be the center ,then ㄥAPO = Θ ,ㄥPOB = 2Θ,
ㄥOPB = (180 - 2Θ) / 2 = 90 - Θ
So ㄥAPB = ㄥAPO + ㄥOPB = Θ + (90 - Θ) = 90
In △APN, PN / a = sinΘ ...(1)
In △BPN, NB / PN = tan(90 - Θ) = 1 / tanΘ...(2)
(1) * (2) : NB / a = sinΘ / tanΘ
NB / a = sinΘ / (sinΘ/cosΘ) = cosΘ
NB = a cosΘ
4) PQ = QR
PN - QN = QN - RN
(ON / tan A) - (ON / tan B) = (ON / tan B) - ( ON / tan X)

(1 / tan A) + (1 / tan X) = 2 / tanB



2009-04-22 19:18:33 補充：
ICorrections of the first Question :

In △APN, PN / a = sinΘ ...(1)

『In △BPN, NB / PN = tan(90 - Θ) = 1 / tanΘ...(2)』

Change to

『In △BPN, NB / PN = tanΘ...(2)』

(1) * (2) : NB / a = sinΘ tanΘ

NB / a = (sinΘ)^2 / cosΘ

NB = a (sinΘ)^2 / cosΘ (ANSWER)

角NPB+角B=90度=角PAB+角B
so 角NPB=角PAB=θ
so 三角形ANP～三角形APB (AA)
so 角APB=角PAN=90度
cosθ=NP/a ,NP=acosθ
sinθ=AN/a ,AN=asinθ
AN*NB=NP^2
so asinθ*NB=a^2*cos^2θ
NB=acos^2θ/sinθ=acosθcotθ

4.
set PQ=QR=a,RN=b,ON=c
2a+b=ccot α---------(1)
a+b=ccot β---------(2)
b=ccotx---------------(3)
2*(2)-(3)
2a+b=2ccotβ-ccotx
so 2ccotβ-ccotx=ccotα
so 2cotβ=cotx+cotα