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?

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

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.

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

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

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'

To find the value of angle APB, we can use the concept of ratios. Since AP:BP:CP=1:2:3, we can assign values to the sides of the square as follows:

Let's assume the side length of the square ABCD is "x". Then, AP = x/6, BP = x/3, and CP = x/2.

Since ABCD is a square, all its angles are right angles (90 degrees). We can use this information to find the value of angle APB.

In triangle APB, we have two known sides, AP and BP. We can use the Law of Cosines to find the angle opposite to the side CP:

cos(APB) = (AP^2 + BP^2 - CP^2) / (2 * AP * BP)

Substituting the values, we have:

cos(APB) = ((x/6)^2 + (x/3)^2 - (x/2)^2) / (2 * (x/6) * (x/3))

cos(APB) = (x^2/36 + x^2/9 - x^2/4) / (2 * x/18)

cos(APB) = (4x^2 + 16x^2 - 9x^2) / (72x)

cos(APB) = 11x^2 / 72x

Simplifying further, we have:

cos(APB) = 11x / 72

To find the value of angle APB, we can substitute the cosine value into the inverse cosine function:

APB = cos^(-1)(11x/72)

Now, since we know that ABCD is a square, the side length "x" is equal to the distance from the point P to any side of the square.

Therefore, the value of angle APB is:

APB = cos^(-1)(11 * CP / 72)

Substituting the value of CP = x/2:

APB = cos^(-1)(11 * (x/2) / 72)

Hence, the value of angle APB in degrees is cos^(-1)(11 * (x/2) / 72).

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.