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Question Number 192469 by Mingma last updated on 18/May/23

Answered by a.lgnaoui last updated on 19/May/23

$$△\mathrm{ABC}\mathrm{triangle}\mathrm{isocele}\mathrm{AB}=\mathrm{AC}$$$$\mathrm{donc}\measuredangle \mathrm{B}=\measuredangle \mathrm{C}=90-\frac{78}{2}=51°$$$$\frac{\sin 78}{\mathbf{BC}}=\frac{\sin 51}{\mathbf{AC}}\left(1\right)$$$$△\mathrm{ABD}\mathrm{AB}=\mathrm{BD}\measuredangle \mathrm{CAD}={\mathrm{A}}_1$$$$$$$$\frac{\sin \mathbf{x}}{\mathbf{AD}}=\frac{\sin (78-{\mathrm{A}}_1)}{\mathbf{BD}}$$$$\mathrm{BDsin}\mathrm{x}=\mathrm{ADsin}(78-{\mathrm{A}}_1)$$$$$$$$\mathbf{BD}=\frac{\mathbf{AD}\sin (78-{\mathbf{A}}_1)}{\sin \mathbf{x}}\left(2\right)$$$$$$$$△\mathrm{ACD}\measuredangle \mathrm{C}=\measuredangle \mathrm{B}=51\measuredangle \mathrm{ADC}=51-30=21$$$$\Rightarrow \frac{\sin 21}{\mathrm{AD}}=\frac{\sin {\mathrm{A}}_1}{\mathrm{C}\mathbf{D}}=\frac{\sin (180-21-{\mathrm{A}}_1)}{\mathrm{AC}}$$$$=\frac{\sin (159-{\mathrm{A}}_1)}{\mathrm{AC}}$$$$\mathbf{AD}\sin {\mathbf{A}}_1=\mathbf{CD}\sin 21\left(3\right)$$$$$$$$\left(1\right)\left(2\right)\Rightarrow$$$$△\mathrm{BCD}\measuredangle \mathrm{DBC}=51-\mathrm{x}$$$$\frac{\sin (51-\mathrm{x})}{\mathrm{CD}}=\frac{\sin 30}{\mathbf{BD}}$$$$$$$$\mathbf{BD}=\frac{\mathbf{CD}\sin 30}{\sin (51-\mathbf{x})}$$$$\left(2\right)\Rightarrow \frac{\mathbf{CD}\sin 30}{\sin (51-\mathbf{x})}=\frac{\mathbf{AD}\sin (78-\mathbf{A})}{\sin \mathbf{x}}$$$$$$$$\mathbf{d}\mathrm{apres}\left(3\right)$$$$\mathbf{AD}\sin \mathbf{A}=\mathbf{CD}\sin 21$$$$\mathbf{CD}=\frac{\mathbf{AD}\sin \mathbf{A}}{\sin 21}\left(5\right)$$$$$$$$\frac{\sin \mathbf{A}\sin 30}{\sin 21.\sin (51-\mathbf{x})}=\frac{\sin (78-\mathbf{A})}{\sin \mathbf{x}}$$$$$$$$$$$$△\mathrm{ABD}\measuredangle \mathrm{BDA}=\measuredangle \mathrm{BAD}=78-{\mathbf{A}}_1$$$$\Rightarrow \mathrm{x}+2(78-{\mathrm{A}}_1)=180$$$${\mathrm{x}}_1+156-{2\mathrm{A}}_1=180{\mathrm{A}}_1=\frac{\mathrm{x}}{2}-12\Rightarrow$$$$78-{\mathrm{A}}_1=90-\frac{\mathbf{x}}{2}\sin (78-{\mathbf{A}}_1)=\cos \frac{\mathbf{x}}{2}$$$$$$$$\mathbf{Equation}\mathbf{est}$$$$$$$$\mathbf{sinx}.\mathbf{cos}(\frac{\mathbf{x}}{2}-12)\mathbf{sin}30=\mathbf{cos}\frac{\mathbf{x}}{2}.\mathbf{sin}(51-\mathbf{x})\mathbf{sin}21$$$$$$$$\frac{1}{2}\sin \mathrm{x}(\cos \frac{\mathrm{x}}{2}\cos 12+\sin \frac{\mathrm{x}}{2}\sin 12=$$$$=\sin 21\left(\cos \frac{\mathrm{x}}{2}\right)(\sin 51.\cos \mathrm{x}-\cos 51\sin \mathrm{x})$$$$\mathrm{diviso}\mathbf{n}\mathrm{s}\mathrm{les}2\mathrm{memvees}\mathrm{pa}\mathbf{r}\cos \frac{\mathrm{x}}{2}$$$$$$$$\frac{1}{2}\sin \mathrm{x}.\cos 12+\tan \frac{\mathrm{x}}{2}\sin 12=$$$$=\sin 21(\sin 51\cos \mathrm{x}-\cos 51\sin \mathrm{x})$$$$$$$$\mathrm{posons}\mathbf{t}=\tan \frac{\mathbf{x}}{2}$$$$\sin \mathbf{x}=\frac{2\mathbf{t}}{1+{\mathbf{t}}^2}\cos \mathbf{x}=\frac{1-{\mathbf{t}}^2}{1+{\mathbf{t}}^2}$$$$\frac{\mathbf{t}}{1+{\mathbf{t}}^2}\cos 12+\mathbf{t}\sin 12=\sin 21(\sin 51\left(\frac{1-{\mathbf{t}}^2}{1+{\mathbf{t}}^2}\right)-\cos 51\frac{2\mathbf{t}}{1+{\mathbf{t}}^2})$$$$\frac{\cos 12.\mathbf{t}}{1+{\mathbf{t}}^2}+\frac{\sin 12(\mathbf{t}(1+{\mathbf{t}}^2)}{1+{\mathbf{t}}^2}=\frac{\sin 21.\sin 51(1-{\mathbf{t}}^2)}{1+{\mathbf{t}}^2}-\frac{2\sin 21\sin 51\mathbf{t}}{1+{\mathbf{t}}^2}$$$$$$$$\frac{\sin 12.{\mathbf{t}}^3+(\sin 21.\sin 51){\mathbf{t}}^2+(\sin 12+\cos 12+2\sin 21\cos 51)\mathbf{t}-\sin 21.\sin 51}{1+{\mathbf{t}}^2}=0$$$$$$$$$$$$\sin 12=0,2079117\cos 12=0,9781476$$$$\sin 51=0,777146\cos 51=0,629320$$$$\sin 21=0,35836$$$$$$$$0,2079117{\mathbf{t}}^3+0,278504{\mathbf{t}}^2+0,663613\mathbf{t}+0,278504=0$$$$$$$$\mathrm{la}\mathrm{re}\mathbf{solutiin}\mathbf{donne}\frac{\mathbf{x}}{2}={\tan }^{-1}(0,734)$$$$\mathrm{soit}\mathbf{x}=72,5°$$$$$$$$$$$$$$

Commented by a.lgnaoui last updated on 19/May/23

Answered by HeferH last updated on 19/May/23

Commented by Mingma last updated on 21/May/23

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