math 3

圖片參考：http://farm5.static.flickr.com/4095/4804122876_c70807331c.jpg



a)ㄥACB = ㄥBAQ = 29° (∠in alt. segment)


ㄥAOB = 2ㄥACB (∠at centre twice ∠ at ☉ce)

= 29 x 2 = 58°


b)ㄥOEC = ㄥACB + ㄥCBO (ext. ∠ of Δ)

By a) ,ㄥACB = 29°

ㄥCBO = ㄥAOB = 58° (alt.∠s, BC//AO)

So ㄥOEC = 29 + 58 = 87°


2010-07-18 16:33:14 補充：
In △BTC , CT / BT = 36/24 = 3/2

In △CTA , AT / CT = (15*2 + 24) / 36 = 3/2

So △BTC ~ △CTA ,

Thus ㄥ CAT = ㄥBCT ,

In △ATC ,

ㄥCAT + ㄥACT + θ = 180 (ㄥ sum of △)

ㄥCAT + (ㄥACB + ㄥBCT) + θ = 180

ㄥCAT + (90 + ㄥCAT) + θ = 180

2ㄥCAT + θ = 90

ㄥCAT = (90 - θ)/2

2010-07-18 16:38:32 補充：
If my answer is correct , please choose me the best.THX!!

2010-07-18 17:31:00 補充：
x^2+y-x=1.....(1)
2x-3y=1.....(2)


(1)*3 + (2) ,

3x^2 + 3y - 3x + 2x - 3y = 1*3 + 1

3x^2 - x = 4

3x^2 - x - 4 = 0

(3x - 4)(x + 1) = 0

x1 = 4/3 or x2 = - 1 ,

sub them to (2) :

2(4/3) - 3y = 1 or 2(-1) - 3y = 1

y1 = 5/9 or y2 = - 1

Exactly, pls. follow the rule here.
If you want to supp. questions, they must be related to the original question you asked, say when you don't understand the ans and want more explanation.

Wing Man, plx don't continually ask supplementary questions...It's unfair for the answerer to answer so many questions...

PLS help to answer...
In the figurre, AB is a diameter of the circle, CT is the tangent to the circle at C, ABT is a straight line. If BT=24 cm, CT=36 cm and angle BTC=θ.
And, radius=15 cm

Express angle CAB in terms of θ.
PLS help, THX A LOT..

2010-07-18 15:49:42 補充：
The diagram of second question :http://www.flickr.com/photos/51470019@N07/4804263350/

THX.

2010-07-18 17:07:14 補充：
OK......

2010-07-18 17:23:12 補充：
PLS answer my last question..
i must give u be the BEST..

x^2+y-x=1
2x-3y=1

PLS solve the above simultaneous equation....
THX a lot..... sorry for ask too many question...