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Solution 1<\/strong>: As shown in the diagram at the right, link $AC$. Without loss of generality, let $AB=BC=CD=DA=x$, $PA=1$, $PB=2$, $PC=3$, $\\angle{APB}=\\alpha$, $\\angle{BPC}=\\beta$. Therefore $AC=\\sqrt{2}x$, and $\\angle{APC}=360^\\circ-(\\alpha+\\beta)$.<\/p>\n\n\n\nApplying the Law of Cosines on $\\triangle{APB}$, $\\triangle{BPC}$, and $\\triangle{CPA}$:<\/p>\n\n\n\n
$$AB^2=PA^2+PB^2-2\\cdot PA\\cdot PB\\cdot cos\\angle{ABP}$$<\/p>\n\n\n\n
$$BC^2=PB^2+PC^2-2\\cdot PB\\cdot PC\\cdot cos\\angle{BPC}$$<\/p>\n\n\n\n
$$AC^2=PA^2+PC^2-2\\cdot PA\\cdot PC\\cdot cos\\angle{CPA}$$<\/p>\n\n\n\n
i.e.<\/p>\n\n\n\n
$$x^2=1^2+2^2-2\\cdot 1\\cdot 2\\cdot cos(\\alpha)$$<\/p>\n\n\n\n
$$x^2=2^2+3^2-2\\cdot 2\\cdot 3\\cdot cos(\\beta)$$<\/p>\n\n\n\n
$$2x^2=1^2+3^2-2\\cdot 1\\cdot 3\\cdot cos(360^\\circ – (\\alpha+\\beta))$$<\/p>\n\n\n\n
Simplifying the above equations<\/p>\n\n\n\n
$$x^2=5-4\\cdot cos(\\alpha)$$<\/p>\n\n\n\n
$$x^2=13-12\\cdot cos(\\beta)$$<\/p>\n\n\n\n
$$x^2=5-3\\cdot cos(\\alpha+\\beta)$$<\/p>\n\n\n\n
Let $cos(\\alpha)=y$, $cos(\\beta)=z$, then $cos(\\alpha+\\beta)=y\\cdot z-\\sqrt{1-y^2}\\cdot\\sqrt{1-z^2}$.<\/p>\n\n\n\n
$$x^2=5-4y$$ $$x^2=13-12z$$ $$x^2=5-3(yz-\\sqrt{1-y^2}\\cdot\\sqrt{1-z^2})$$<\/p>\n\n\n\n
Replacing $x^2$ with $5-4y$, $z$ with $\\dfrac{y+2}{3}$ in the above:<\/p>\n\n\n\n
$$5-4y=5-3(y\\cdot\\dfrac{y+2}{3}-\\sqrt{1-y^2}\\cdot\\sqrt{1-(\\dfrac{y+2}{3})^2})$$<\/p>\n\n\n\n
Simplifying and factorizing the above: $$(4y-5)(2y^2-1)=0$$ Solving the above equation, $$y=\\dfrac{5}{4}\\ \\ \\ \\ \\ or \\ \\ \\ \\ \\ y=-\\dfrac{\\sqrt{2}}{2}$$ Ignoring the invalid value, we have $cos(\\alpha)=-\\dfrac{\\sqrt{2}}{2}$. Thus $$\\angle{APB}=\\alpha=\\boxed{135^\\circ}$$<\/p>\n\n\n\n
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