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Question Number 183246 by cortano1 last updated on 24/Dec/22

Answered by mr W last updated on 24/Dec/22

Commented by mr W last updated on 24/Dec/22

$$\cos \alpha =\frac{1^2+{\left(2\sqrt{2}\right)}^2-3^2}{2\times 1\times 2\sqrt{2}}=0\Rightarrow \alpha =90°$$$$\Rightarrow x=\alpha +45°=135°$$

Commented by cortano1 last updated on 24/Dec/22

$$nice$$

Answered by HeferH last updated on 24/Dec/22

Commented by HeferH last updated on 24/Dec/22

$$3^2=1^2+{\left(2\sqrt{2}\right)}^2\Rightarrow righttriangle$$$$\alpha -45°=90°$$$$\alpha =135°$$

Commented by cortano1 last updated on 24/Dec/22

$$great$$

Answered by SEKRET last updated on 04/Jan/23

$$\mathbf{ABP}=\mathbf{a}\mathbf{PBC}=90-\mathbf{a}$$$$\mathbf{AB}=\mathbf{BC}=\mathbf{x}$$$${\mathbf{x}}^2+4-4\mathbf{xcos}\left(\mathbf{a}\right)=1\mathbf{cos}\left(\mathbf{a}\right)=\frac{{\mathbf{x}}^2+3}{4\mathbf{x}}$$$${\mathbf{x}}^2+4-4\mathbf{xsin}\left(\mathbf{a}\right)=9\mathbf{sin}\left(\mathbf{a}\right)=\frac{{\mathbf{x}}^2-5}{4\mathbf{x}}$$$${\mathbf{sin}}^2\left(\mathbf{a}\right)+{\mathbf{cos}}^2\left(\mathbf{a}\right)=1$$$${({\mathbf{x}}^2+3)}^2+{({\mathbf{x}}^2-5)}^2=16{\mathbf{x}}^2$$$$2{\mathbf{x}}^4-4{\mathbf{x}}^2+34-16{\mathbf{x}}^2=$$$$\mathbf{x}=\sqrt{5+2\sqrt{2}}$$$$1^2+2^2-4\mathbf{cos}\left(\mathbf{b}\right)=5+2\sqrt{2}$$$$\mathbf{cos}\left(\mathbf{b}\right)=\frac{-\sqrt{2}}{2}\mathbf{b}=135°$$$$$$$$$$

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