kimsiang
18-10-2005, 06:50 PM
Help!
Points A and C have coordinates (-1,2) and (9,7) respectively. A rectangle ABCD has AC as a diagonal. Calculate the possible coordinates of B and D if the length of AB is 10 units
Here is my solution.....but it is quite long...shd hv other shorter solution i think.......
Since AC is a diagonal , so do BD.
Hence AC = BD = 10 units (#)
since the coordinates of A,C were given....AC^2 = 125 units^2
By using the fact that AC is the hypotenus of triangles ABC and also ADC,including (#),we can form 4 equations.
First,let coordinate B = (b1,b2) , and coordinate D = (d1,d2)
then we wil have
(b1 - 9)^2 + (b2 - 7)^2 = 25 (1) BC = 5 units
(b1 +1)^2 + (b2 - 2)^2 = 100 (2) AB = 10 units
(d1 +1)^2 + (d2 - 2)^2 = 25 (3) AD = 5 units
(d1 - 9)^2 + (d2 - 7)^2 = 100 (4) DC = 10 units
solving (1) ,(2) simultanously and then (3),(4),you can get
2b1 + b2 = 20
2d1 + d2 = 5
Substitute b1 or b2,d1 or d2 back to either (1) or (2) and (3) or (4),
you will get 2 quadratic equations.....
For example,substitute d2 = 5 - 2d1 into (3),
we can get (d1 +1)^2 + ([5 - 2d1] - 2)^2 = 25
(d1 +1)^2 + (3 - 2d1)^2 = 25
.
.
.
d1^2 - 2d1 - 3 =0
Hence d1 = -1 or 3
d2 = 7 or -1
then we can have D(-1 ,7) , D(3,-1).
For b1,b2,under the same process,you shd get the answer too.
It is b1 = 5 or 9
b2 = 10 or 2
then we can have B(5,10) , B(9,2).
Now we have to find which B is corespond to which D.
since both AC and BD are diagonals,the shd have the same center point,that is (4,9/2).We can get the sets we want.
Hence,the possible coordinate sets for B and D are
B(5,10) and D(3,-1) , B(9,2) and D(-1 ,7) .
________
HALF-BAKED (div>
1) sin x - sin 3x + sin 2x = 2 sin x
1 + cos x ? 2cos2 x
2) sin 3x + sin 6x = sin 9x
does anybody can help me to solve this question ?
thank you a lot!!
there is another question
prove that the area of any rhombus in given by A = 1/2 xy where x and y are the lengths of the diagonals. [8]
thank you !!
kimsiang
26-10-2005, 06:42 PM
1) sin x - sin 3x + sin 2x = 2 sin x
1 + cos x ? 2cos2 x
2) sin 3x + sin 6x = sin 9x
3)prove that the area of any rhombus is given by A = 1/2 xy where x and y are the lengths of the diagonals.
Solution)
I think there should be domain for x,for example, -180 =< x =< 180..
so,since i duno the range,i wil use -180 =< x = < 180 here...
1.when x=0 , 1 + cos(x) ? 2cos(2x)= infinity,so,x not = 0.
when x not = 0, sin(x)-sin(3x)+sin(2x)=2sin(x)[1+cos(x)?2cos(2x)]
<=> sin(x)-sin(3x)+sin(2x) =2sin(x)+2sin(x)cos(x)?4sin(x)cos(2x)
<=> -sin(3x)+sin(2x) = sin(x)+sin(2x) ?4sin(x)cos(2x)
<=> -sin(3x) = sin(x) ?4sin(x)cos(2x)
<=> 4sin(x)cos(2x) = sin(x)+sin(3x)
<=> 4sin(x)[2cos(x)^2-1] = 2sin(2x)cos(x)
<=>8sin(x)cos(x)^2-4sin(x)=4sin(x)cos(x)^2
<=> 4sin(x)cos(x)^2 - 4sin(x) = 0
<=> sin(x)[cos(x)^2 - 1] = 0
<=> sin(x) = 0 or cos(x)^2 - 1 = 0
<=> x = -180,180 or cos(x) = -1 , 1
<=> x = -180,180 or x = -180,180
so,x = -180,180
2. sin(3x) + sin(6x) = sin(9x)
Obviously, x =-180,0,180 are 3 of the solution(s).
When x not =-180,0,180,
<=>sin(3x)+2sin(3x)cos(3x)=sin(3x)cos(6x)+sin(6x)cos(3x)
<=>sin(3x)[1+2cos(3x)]=sin(3x)[2cos(3x)^2-1]+2sin(3x)cos(3x)^2
since x not =-180,0,180,sin(3x) not = 0,then we can cancel it from both sides.
<=> 1+2cos(3x)=2cos(3x)^2-1+2cos(3x)^2
<=> 0=4cos(3x)^2-2cos(3x)-2
<=> 2cos(3x)^2-cos(3x)-1=0
<=> cos(3x)= [1/4] (+/-) [sqrt( 1 + 8 )/4]
<=> cos(3x)= [1/4] (+/-) [3/4]
<=> cos(3x)= -1/2 , 1
Since -180 =< x = < 180 , -540 =< 3x = < 540
<=> 3x= -480,-240,-120,120,240,480,360,-360
<=> x= -160,-120,-80,-40,40,80,120,160
So,the solutions r x=-180,-160,-120,-80,-40,0,40,80,120,160,180.
3.
images/eps-gif/Rhombus_750.gif
By refering to the diagram above,
lets devide the area into 4 parts(4 right triangles)
Area of 1 right triangle = 0.5*(base*height)
= 0.5*(0.5p*0.5q) = 0.125pq
Since there r 4 right triangles, TOtal area = Area of Rhombus
= 0.125pq * 4
= 0.5pq
________
LovelyWendie99 ()
ops... sorry for my careless
actually the questions is requiring proving
2) prove that sin 3x + sin 6x = sin 9x
sorry and thank you for ur help, kimsiang!
kimsiang
27-10-2005, 11:28 AM
2) prove that sin 3x + sin 6x = sin 9x
sorry and thank you for ur help, kimsiang!
But sin 3x + sin 6x = sin 9x is not an equation...
for example,when x = 15 ,
sin 3x + sin 6x = sin 45 + sin 90 = (sqrt 2)/2 + 1
but sin 9x = sin 135 = sin 45 = (sqrt 2)/2
means (sin 3x + sin 6x) not = sin 9x
So,the original question that u have typed shd be ok..
________
THE CIGAR BOSS ()
confused-freaker
28-10-2005, 10:51 AM
Hi there, i got some problems with this q:
11. (b) A large quantity of shoes from brand B is inspected before been distributed to slaes outlets. It is found that 1% of the left side shoes are defective while 2% of the right side shoes are defective. By assuming that the defects occur independently and using suitable approximation, calculate the probability that in a random sample of 50 pairs of shoes,
(i) there are two pairs of shoes with defects on their left side.
(ii) there is a pair of shoes with defects on both sides.
[stpm 1999]
answers from the 5-year series:
11. (b) (i) 0.076 (ii) 0.251
kimsiang
28-10-2005, 06:11 PM
Hi there, i got some problems with this q:
11. (b) A large quantity of shoes from brand B is inspected before been distributed to slaes outlets. It is found that 1% of the left side shoes are defective while 2% of the right side shoes are defective. By assuming that the defects occur independently and using suitable approximation, calculate the probability that in a random sample of 50 pairs of shoes,
(i) there are two pairs of shoes with defects on their left side.
(ii) there is a pair of shoes with defects on both sides.
[stpm 1999]
(i)For left side shoes : n=50,p(defective)=0.01 => np=0.5...then we can use poisson distribution as approximation...
P(X=2)=[e^(-0.5)]*[(0.5)^2]*/2 = 0.075816332
= 0.076
(#) Here,n means number of left side shoe,X means number of left side shoe which is(are) defective.
(ii)Lets consider each pair of shoes...
p(both defective)=0.01*0.02=0.0002
p(not both defective)= 1 - 0.0002 = 0.9998 (or u can add up 0.01*0.98,0.99*0.02,0.98*0.99)
m=50,p=0.0002 =>mp=0.01
Poisson again......
P(Y=1)=[e^(-0.01)]*[0.01] = 0.0099
(#) Here,m means number of pairs,Y means number of pairs which is(are) both defective.
There r maybe other ways to solve (ii)......maybe there r better n more accurate solution....but this is wat i can get
since i din study that for a long time...Btw,The answer given is definitely strange....dun u feel that?
________
LovelyWendie99 ()
confused-freaker
28-10-2005, 08:36 PM
Yea, the part (i) i can do as normal....comes to part (ii)....i'm kinda blur why their working diff d........let me type out the whole thing for you:
Let X be the number of pairs of brand B shoes, out of 50 pairs, which have defects on the left sides, and that Y the number of pairs which have defects on the right side, then
X~B(50,0.01)
Y~B(50, 0.02)
(i) Since n =50 is large and np = 0.5 < 5, and so
X~P(0.5) approximately,
P(X=2) = e^(-0.5) * 0.5^2 / 2! = 0.0758 (this is ok)
(ii) Since n = 50 is large and np = 1 < 5, hence
Y~P(1) approximately
Z = X + Y
so Z~P(1.5)
P(Z=2) = e^(-1.5) * 1.5^2 / 2! = 0.251
I dont really get why they go and plus them...never consider the 0.01x0.02 thing......sigh
Anyway, if i dont use this poisson approx, can i use the binomial approx?
confused-freaker
29-10-2005, 12:39 AM
another q:
Given two parallel lines l1 and l2 passing through (5, 0) and (-5, 0) respectively, and meet the line 4x + 3y = 25 respectively at P and Q. If PQ equals to 5 units, find the possible slopes of l1 and l2.
[stpm 2001 p1 q5]
sorry for flooding so many q =.= but thanks in advance.
The answer/working uses the angle n tangent thing....but is there any other easier to understand way?
kimsiang
29-10-2005, 03:30 PM
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