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Question Number 208385 by efronzo1 last updated on 14/Jun/24

Commented by efronzo1 last updated on 15/Jun/24

Commented by efronzo1 last updated on 15/Jun/24

$${\mathrm{i}}^{\prime }\mathrm{m}\mathrm{stuck}\mathrm{this}\mathrm{step}$$

Commented by mr W last updated on 15/Jun/24

$$\Rightarrow y=\frac{2+x}{3}$$$$\Rightarrow z=\frac{4x}{3}$$$$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sqrt{(1-{\cos }^2\alpha )(1-{\cos }^2\beta )}$$$$z=xy-\sqrt{(1-x^2)(1-y^2)}$$$${(xy-z)}^2=(1-x^2)(1-y^2)$$$$x^2{y}^2+z^2-2xyz=1-x^2-y^2+x^2{y}^2$$$$z^2-2xyz=1-x^2-y^2$$$$\frac{16x^2}{9}-\frac{8x^2(2+x)}{9}=1-x^2-\frac{{(2+x)}^2}{9}$$$$8x^3-10x^2-4x+5=0$$$$\Rightarrow x=\pm \frac{\sqrt{2}}{2},\frac{5}{4}$$$$onlyx=-\frac{\sqrt{2}}{2}issuitable$$$$\cos \alpha =x=-\frac{\sqrt{2}}{2}\Rightarrow \alpha =135°$$

Answered by mr W last updated on 14/Jun/24

$$s^2=\frac{p^2+r^2+\sqrt{4(p^2+r^2-q^2)q^2-{(r^2-p^2)}^2}}{2}$$$$=\frac{1^2+3^2+\sqrt{4(1^2+3^2-2^2)2^2-{(3^2-1^2)}^2}}{2}$$$$=5+2\sqrt{2}$$$$\cos \mathrm{\angle }APB=\frac{1^2+2^2-(5+2\sqrt{2})}{2\times 1\times 2}=-\frac{\sqrt{2}}{2}$$$$\Rightarrow \mathrm{\angle }APB=135°$$

Commented by efronzo1 last updated on 14/Jun/24

$$\mathrm{what}$$

Commented by mr W last updated on 15/Jun/24

Commented by mr W last updated on 15/Jun/24

$$\cos \alpha =\frac{s^2+q^2-p^2}{2qs}$$$$\sin \alpha =\cos \beta =\frac{s^2+q^2-r^2}{2qs}$$$${\left(\frac{s^2+q^2-p^2}{2qs}\right)}^2+{\left(\frac{s^2+q^2-r^2}{2qs}\right)}^2=1$$$${(s^2+q^2-p^2)}^2+{(s^2+q^2-r^2)}^2=4q^2{s}^2$$$$2s^4-2(p^2+r^2)s^2+p^4+r^4+2q^4-2(p^2+r^2)q^2=0$$$$\Rightarrow s^2=\frac{p^2+r^2\pm \sqrt{4(p^2+r^2-q^2)q^2-{(r^2-p^2)}^2}}{2}$$$$+forpointPinsidethesquare$$$$-forpointPoutsidethesquare$$

Commented by Tawa11 last updated on 21/Jun/24

$$\mathrm{Weldone}\mathrm{sir}.$$

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