在正方形中找角度（用畢氏定理）問題分數:10marks

let A is (0,0) , B is (0,a) , C is (a,0) , P is (x,y)

then we get :

x^2+y^2=1 --(1)
x^2+y^2-2ay+a^2-4=0--(2)
x^2+y^2-2ax+a^2-9=0--(3)

solve the equation , we get
a=2.982937438
x=0.150508648
y=0.988608692

....then angle APB=167.0277599.

Let side of the square ABCD=n, therefore, AB=AC=n.
Let AngleBAP=x.
For triangle BAP, using cosine rule:
2^2 = n^2 +1^2 - 2(1)(n)cosx..........................(1)
For triangle PAC again using cosine rule:
3^2= 1^2 + n^2 -2(1)(n)cos(90-x) (since angle BAC is a rt-angle.)
or 3^2 = 1^2 +n^2 -2(1)(n)sinx........................(2)
From (1), we get 2ncosx =n^2-3................(3)
From (2), we get 2nsinx = n^2 -8................(4)
Therefore, (2ncosx)^2 + (2nsinx)^2 = (n^2-3)^2 + (n^2 -8)^2
4n^2 = n^4 -6n^2 +9 + n^4 -16n^2 + 64
2n^4 -26n^2 +73 = 0
Solving the quadratic equation we get n^2=8.898 or 4.102, that is side of the square n=2.983 or 2.025(reject because it cannot accomodate PC=3)
Again applying cosine rule to triangleBAP, therefore,
n^2 = 2^2 + 1^2 -2(2)(1) cos(angle APB)=8.898
therefore, angle APB=cos^-1[(5-8.898)/4] = 167 degree.

Rotating triangle PAC about point A through 270 º clockwise, we get C coincide with B and P goes to Q.

Now, draw the line PQ and we have

AP＝AQ＝1
∠QAP＝∠BAP＋∠QAB＝∠BAP＋∠PAC＝∠BAC＝90 º

Therefore,

PQ＝√(1²＋1²)＝√2
∠APQ＝(180－90)/2＝45 º

By the cosine law, we have

BQ²＝BP²＋PQ²－2BP．PQcos(∠QPB)
＝＞ 9＝4＋2－2．2√2cos(∠QPB)
＝＞ cos(∠QPB)＝－3/(4√2)
＝＞ ∠QPB＝122.03 º

∠APB＝∠APQ＋∠QPB＝45＋122.03＝167.03 º

－－－－－－－－－－－－－－－－－－－－－－－－－－－－－

However, from your diagram, I think you mean PD=3 instead of PC=3. In such a case, consider rotating triangle BPD about point B through 270 º counterclockwise, we get D coincide with A and P goes to Q.

Now, draw the line PQ and we have

BP＝BQ＝2
∠PBQ＝∠PBA＋∠ABQ＝∠PBA＋∠DBP＝∠DBA＝90 º

Therefore,

PQ＝√(2²＋2²)＝√8
∠QPB)＝(180－90)/2＝45 º

As AP²＋PQ²＝1＋8＝9＝AQ², by the Pythagorean theorem, we have

∠APQ＝90 º
∠APB＝∠APQ＋∠QPB＝90＋45＝135 º