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Question Number 195043 by necx122 last updated on 22/Jul/23

Commented by necx122 last updated on 22/Jul/23

$$Please,Imeedhelpwiththis.{It}^{\prime }squite$$$$trickyformetotackle.$$

Answered by a.lgnaoui last updated on 23/Jul/23

$$\bullet \mathbf{C}\mathrm{ote}\mathbf{AB}=\mathbf{a}+\mathbf{b}$$$$\{\begin{array}{l}\sin \mathit{\alpha }=\frac{2}{\mathrm{a}}\Rightarrow \mathbf{a}=\frac{2}{\sin \mathit{\alpha }}\\ \cos \mathit{\alpha }=\frac{2}{\mathrm{b}}\Rightarrow \mathbf{b}=\frac{2}{\cos \mathit{\alpha }}\end{array}$$$$\mathbf{a}+\mathbf{b}=2\left(\frac{\sin \mathit{\alpha }+\cos \alpha }{\sin \mathit{\alpha }\cos \mathit{\alpha }}\right)$$$$$$$$\bullet \mathbf{Cote}\mathbf{AC}=\mathbf{c}+\mathbf{d}+\mathbf{e}$$$$\{\begin{array}{l}\sin (90-\mathit{\alpha })=\frac{3}{\mathrm{c}}\Rightarrow \mathbf{c}=\frac{3}{\cos \mathit{\alpha }}\\ \sin (90-\mathit{\alpha })=\frac{2}{\mathbf{d}}=\frac{1}{\mathbf{e}}\end{array}$$$$\Rightarrow \mathbf{c}=\frac{3}{\cos \mathit{\alpha }};\mathbf{d}=\frac{2}{\cos \mathit{\alpha }};\mathbf{e}=\frac{\mathbf{d}}{2}$$$$\mathbf{alors}:\mathbf{AC}=\mathbf{c}+\mathbf{d}+\mathbf{d}=\frac{6}{\cos \mathit{\alpha }}$$$$\mathbf{AB}=\mathbf{AC}(\mathbf{quadrilatere}=\mathbf{care})$$$$\Rightarrow \frac{6}{\cos \mathit{\alpha }}=\frac{2(\sin \mathit{\alpha }+\cos \mathit{\alpha }}{\sin \mathit{\alpha }\cos \mathit{\alpha }}$$$$4\sin \mathit{\alpha }\cos \mathit{\alpha }=\sin \mathit{\alpha }\cos \mathit{\alpha }+\cos {}^2\mathit{\alpha }$$$$4=1+\frac{\cos \mathit{\alpha }}{\sin \mathit{\alpha }}\Rightarrow \tan \mathit{\alpha }=\frac{1}{3}$$$$\mathit{\alpha }=18,43$$$$\mathbf{donc}\mathbf{AB}=\mathbf{AC}=\frac{6}{\cos \mathit{\alpha }}=6\sqrt{1+\tan {}^2\mathit{\alpha }}$$$$\mathbf{cote}\mathbf{du}\mathbf{care}=6\sqrt{1+\frac{1}{9}}=2\sqrt{10}$$$$$$$$\Rightarrow \mathbf{Aire}\mathbf{du}\mathbf{Care}=40$$$$$$

Commented by a.lgnaoui last updated on 23/Jul/23

Commented by JDamian last updated on 23/Jul/23

$$\mathrm{I}\mathrm{get}45\mathrm{without}\mathrm{trigonometric}\mathrm{expressions}$$

Commented by JDamian last updated on 23/Jul/23

$$\frac{6}{\cancel{\cos \mathit{\alpha }}}=\frac{2(\sin \mathit{\alpha }+\cos \mathit{\alpha })}{\sin \mathit{\alpha }\cancel{\cos \mathit{\alpha }}}$$$$3=\frac{\sin \mathit{\alpha }+\cos \mathit{\alpha }}{\sin \mathit{\alpha }}=1+\frac{\cos \mathit{\alpha }}{\sin \mathit{\alpha }}$$$$2=\frac{\cos \mathit{\alpha }}{\sin \mathit{\alpha }}$$$$\tan \mathit{\alpha }=\frac{1}{2}$$$$$$$$\mathbf{donc}\mathbf{AB}=\mathbf{AC}=6\sqrt{1+\tan {}^2\mathit{\alpha }}=$$$$=6\sqrt{1+{\left(\frac{1}{2}\right)}^2}=6\sqrt{\frac{5}{4}}=3\sqrt{5}$$

Commented by a.lgnaoui last updated on 23/Jul/23

$$\mathrm{correction}\left(\mathrm{erreur}\mathrm{calcul}\right)$$$$3=1+\frac{\cos \alpha }{\sin \alpha }\Rightarrow \tan \alpha =\frac{1}{2}$$$$\mathrm{Cote}=\frac{6}{\cos \alpha }=6\sqrt{1+\tan {}^2\alpha }=3\sqrt{5}$$$$\mathrm{Aire}=45$$$$$$

Commented by necx122 last updated on 23/Jul/23

$$Alsoclear.$$

Answered by mr W last updated on 23/Jul/23

Commented by mr W last updated on 23/Jul/23

$$\tan \theta =\frac{2}{3-1}=2$$$$\frac{6}{a}=\sin \theta =\frac{2}{\sqrt{2^2+1^2}}=\frac{2}{\sqrt{5}}$$$$\Rightarrow a=3\sqrt{5}$$$$areaofsquare=a^2={\left(3\sqrt{5}\right)}^2=45$$

Commented by necx122 last updated on 23/Jul/23

$$Thankyousomuch.Itsveryclear.$$

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