ABCD is a square. P is a point inside the square, such that AP:BP:CP=1:2:3. What is the value (in degree) of angle APB?
9 answers
Plz give the full worked solution. I have failed find the answer.
Plz give the full worked solution.I have failed to find the answer.
well, we know that if AP=1, BP=2,CP=3, the side s of the square is 2<s<3.
So, if P = (x,y), we have
x^2+y^2 = 1
(x-s)^2+y^2 = 4
(x-s)^2+(y-s)^2 = 9
Solve those in your favorite way, and you come up with
P = (0.86,0.51) and the square has side
s = 2.80
So, using the law of cosines,
2.80^2 = 1+4-4cos(APB)
APB = 135.2 degrees
So, if P = (x,y), we have
x^2+y^2 = 1
(x-s)^2+y^2 = 4
(x-s)^2+(y-s)^2 = 9
Solve those in your favorite way, and you come up with
P = (0.86,0.51) and the square has side
s = 2.80
So, using the law of cosines,
2.80^2 = 1+4-4cos(APB)
APB = 135.2 degrees
First draw the square ABCD and indicate a point P inside the square near the side AB about 1/3 from A to B.
Label the lengths AP=1, PB=2, PD=3, and each side of the square s.
Join BD and mark the length (√2)s.
Label ∠APB as α, ∠DPA as β.
Consider each triangle APB, BPD and DPA in turn, and write an equation using the cosine rule:
s² = 1²+2²-2(1)(2)cos(α) ....(1)
2s² = 2²+3²-2(2)(3)cos(2π-α-β) ....(2)
s²=3²+1²-2(3)(1)cos(β) ....(3)
From the 3 equations above, eliminate s and substitute a,b for α, β:
f1(a,b)=4cos(a)+6cos(b)-12cos(a+b)-2=0
f2(a,b)=6*cos(b)-4*cos(a)-5=0
[note that cos(2π-a-b) ≡ cos(a+b) ]
Solve the non-linear system by iteration or Newton's method to get:
a=2.915184356217331
b=1.386066761845012
which when converted to degrees, give:
a=α=167.0277601°
b=β=79.4157756°
approximately.
Check:
f1(a,b)=-8.9*10E-15
f2(a,b)=-8.9*10E-16 ok.
Label the lengths AP=1, PB=2, PD=3, and each side of the square s.
Join BD and mark the length (√2)s.
Label ∠APB as α, ∠DPA as β.
Consider each triangle APB, BPD and DPA in turn, and write an equation using the cosine rule:
s² = 1²+2²-2(1)(2)cos(α) ....(1)
2s² = 2²+3²-2(2)(3)cos(2π-α-β) ....(2)
s²=3²+1²-2(3)(1)cos(β) ....(3)
From the 3 equations above, eliminate s and substitute a,b for α, β:
f1(a,b)=4cos(a)+6cos(b)-12cos(a+b)-2=0
f2(a,b)=6*cos(b)-4*cos(a)-5=0
[note that cos(2π-a-b) ≡ cos(a+b) ]
Solve the non-linear system by iteration or Newton's method to get:
a=2.915184356217331
b=1.386066761845012
which when converted to degrees, give:
a=α=167.0277601°
b=β=79.4157756°
approximately.
Check:
f1(a,b)=-8.9*10E-15
f2(a,b)=-8.9*10E-16 ok.
Steve, how did you solve that three equations? And according to your solution, where is the origine?
Mathmate, it seems that, you took DP=3, but according to my question CP=3. So, you have drawn BD, but it should be CD. Though I can not understand the process of iteration or Newton's method to solve that two equations. Can you please clear that part?
I am sorry about the incorrect designations.
However, the same process can be used to solve for CP=3, although the result would be different.
It seems to me that Steve's method is simpler, so use his method.
Using Newton's method to solve for a non-linear equation f(x)=0 with a single unknown is well-known iterative method, namely:
xi+1=xi-f(xi)/f'(xi)
It has its problems of convergence, but most of the time when the initial values are close enough to the solution (within established criteria), it converges very rapidly.
The same method applied to system of non-linear equations is equally useful. It is a similar equation as above, except that x becomes a vector X, and
1/f'(xi) is replaced by the inverse of the Jacobian.
So
Xi+1=Xi-J-1 * f(xi)
For more information, read:
http://laser.cheng.cam.ac.uk/wiki/images/8/8b/NumMeth_Handout_5.pdf
or google
newton's method for non-linear systems
However, the same process can be used to solve for CP=3, although the result would be different.
It seems to me that Steve's method is simpler, so use his method.
Using Newton's method to solve for a non-linear equation f(x)=0 with a single unknown is well-known iterative method, namely:
xi+1=xi-f(xi)/f'(xi)
It has its problems of convergence, but most of the time when the initial values are close enough to the solution (within established criteria), it converges very rapidly.
The same method applied to system of non-linear equations is equally useful. It is a similar equation as above, except that x becomes a vector X, and
1/f'(xi) is replaced by the inverse of the Jacobian.
So
Xi+1=Xi-J-1 * f(xi)
For more information, read:
http://laser.cheng.cam.ac.uk/wiki/images/8/8b/NumMeth_Handout_5.pdf
or google
newton's method for non-linear systems
Even a little thought would reveal that A=(0,0) and C=(s,s).
(s-x)^2 + y^2 = 4
s^2 - 2sx + y^2 = 4
But x^2+y^2 = 1, so
s^2 - 2sx = 3
(s-x)^2 + (s-y)^2 = 9
s^2-2sx+x^2 + s^2-2sy+y^2 = 9
From above, that means
3+1 + s^2-2sy = 9
s^2-2sy = 5
So, now we know that
x = (s^2-3)/2s
y = (s^2-5)/2s
Going back to the first equation, then:
[(s^2-3)/2s)^2 + [(s^2-5)/2s)^2 = 1
s^4-6s^2+9 + s^4-10s^2+25 = 4s^2
2s^2-20s+34 = 0
s^2-10s+17 = 0
s^2 = 5±2√2
The only value for s such that 2<s<3 is
s = √(5+2√2)
s = 2.7979 or 2.80
and x,y falls out from there
(s-x)^2 + y^2 = 4
s^2 - 2sx + y^2 = 4
But x^2+y^2 = 1, so
s^2 - 2sx = 3
(s-x)^2 + (s-y)^2 = 9
s^2-2sx+x^2 + s^2-2sy+y^2 = 9
From above, that means
3+1 + s^2-2sy = 9
s^2-2sy = 5
So, now we know that
x = (s^2-3)/2s
y = (s^2-5)/2s
Going back to the first equation, then:
[(s^2-3)/2s)^2 + [(s^2-5)/2s)^2 = 1
s^4-6s^2+9 + s^4-10s^2+25 = 4s^2
2s^2-20s+34 = 0
s^2-10s+17 = 0
s^2 = 5±2√2
The only value for s such that 2<s<3 is
s = √(5+2√2)
s = 2.7979 or 2.80
and x,y falls out from there
Yeah...got it. And I have also find a solution, without co-ordinate geometry...if we take the point 'B' at the bottom right side of the figer, and the point 'A' in the bottom left side, then, rotate the triangle BPC 90 degrees in left side, with respect to 'B'. Then, join A , P'
1 answer