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Question Number 202938 by mr W last updated on 06/Jan/24

Commented by esmaeil last updated on 06/Jan/24

$$18.4349$$$$?$$$$$$

Commented by mr W last updated on 06/Jan/24

$$no.$$

Commented by mr W last updated on 11/Jan/24

Answered by a.lgnaoui last updated on 11/Jan/24

$$△\mathbf{FBN}\measuredangle \mathrm{b}+90+60+\measuredangle {\mathrm{c}}_1=180$$$$\measuredangle {\mathrm{c}}_1=30-\measuredangle \mathrm{b}\mathrm{BC}=\mathrm{q}\mathrm{BF}=\mathrm{EF}=\mathrm{p}$$$$\frac{\sin \mathrm{b}}{\mathrm{q}}=\frac{\sin (30-\mathrm{b})}{\mathrm{p}}\left(1\right)$$$$△\mathbf{AFB}\frac{\sin (45+\mathrm{b})}{\mathrm{q}}=\frac{\mathrm{sina}}{\mathrm{p}}\left(2\right)$$$$\left(1\right)\mathrm{et}\left(2\right)\frac{\sin (30-\mathrm{b})}{\sin \mathrm{b}}=\frac{\mathrm{sina}}{\sin (45+\mathrm{b})}$$$$\mathrm{a}+\mathrm{b}=45\Rightarrow \mathrm{a}=45-\mathrm{b}$$$$$$$$\mathrm{donc}\frac{\sin (30-\mathrm{b})}{\sin \mathrm{b}}=\frac{\sin (45-\mathrm{b}}{\sin (45+\mathrm{b})}$$$$\frac{(1-\sqrt{3}\tan \mathrm{b})}{2\tan \mathrm{b}}=\frac{1-\tan \mathrm{b}}{1+\tan \mathrm{b}}$$$$\Rightarrow (2-\sqrt{3})\tan {}^2\mathbf{b}-(1+\sqrt{3})\tan \mathbf{b}+1=0$$$$\Rightarrow \mathbf{b}=20,81$$$$\mathrm{donc}\mathbf{a}=24,19$$$$\mathrm{Calcul}\mathrm{de}\mathbf{x}$$$$△\mathrm{AEF}\frac{\sin \mathbf{x}}{\mathrm{EF}}=\frac{\sin \mathbf{a}}{\mathrm{AE}}$$$$\frac{\sin \mathbf{x}}{\mathbf{p}}=\frac{\sin \mathbf{a}}{\mathrm{AE}}$$$$△\mathrm{ADE}\frac{\sin (\mathbf{x}+\mathbf{a})}{\mathbf{p}}=\frac{1}{\mathrm{AE}}$$$$\Rightarrow \mathrm{AE}=\frac{\mathbf{p}}{\sin (\mathbf{x}+\mathbf{a})}$$$$\frac{\sin \mathbf{x}}{\mathbf{p}}=\frac{\sin \mathbf{a}\times \sin (\mathbf{x}+\mathbf{a})}{\mathbf{p}}$$$$\sin \mathbf{x}=\sin \mathbf{a}(\sin \mathbf{x}\cos \mathbf{a}+\cos \mathbf{x}\sin \mathbf{a})$$$$=\sin \mathbf{a}\times \cos \mathbf{x}(\tan \mathbf{x}\cos \mathbf{a}+\sin \mathbf{a})$$$$\tan \mathbf{x}=\sin \mathbf{a}(\tan \mathbf{x}\cos \mathbf{a}+\sin \mathbf{a})$$$$\mathbf{posons}\mathbf{t}=\tan \mathbf{x}$$$$\mathbf{t}=\sin \mathbf{a}\cos \mathbf{a}\times \mathbf{t}+\sin {}^2\mathbf{a}$$$$\mathbf{t}=\frac{\sin {}^2\mathbf{a}}{1-\sin \mathbf{a}\cos \mathbf{a}}=0,2681282$$$$$$$$\mathbf{Soit}\mathbf{x}=15°$$$$$$

Commented by mr W last updated on 11/Jan/24

$$15°iscorrect.$$

Commented by a.lgnaoui last updated on 11/Jan/24

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