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Question Number 202056 by mr W last updated on 19/Dec/23

Commented by mr W last updated on 19/Dec/23

$$thelakewithcenteratOhasaradius$$$$r(r=1km).theshortestdistances$$$$fromthevillagesAandBtothe$$$$lakeareaandbrespectively(a=4km,$$$$b=2km).apumpstationshouldbe$$$$buildatCtosupplythevillages$$$$withwater.findtheminimum$$$$lengthofpipesforthisproject.$$$$(\theta =60°).$$

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

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

$$\mathrm{\Delta }\mathbf{OAB}$$$${\mathbf{AB}}^2={\mathbf{OA}}^2+{\mathbf{OB}}^2-2\mathbf{OA}.\mathbf{OB}\cos 60$$$$=25+9-15=34-15=19$$$$\Rightarrow \mathbf{AB}=\sqrt{19}$$$${\mathbf{AB}}^2={\mathbf{AC}}^2+{\mathbf{CB}}^2$$$$(\measuredangle \mathbf{ACB}=90)\left[\mathrm{Courte}\mathrm{distance}\mathrm{entre}\mathbf{A}\mathbf{B}\right]$$$$\mathrm{\Delta }\mathbf{OAC}\measuredangle \mathrm{DOC}=\mathbf{x}$$$${\mathbf{AC}}^2={\mathbf{OC}}^2+{\mathbf{AD}}^2-2.\mathbf{OC}.\mathbf{AD}\cos \mathbf{x}$$$$=26-10\cos \mathbf{x}$$$$\Rightarrow \mathbf{AC}=\sqrt{26-10\cos \mathbf{x}}$$$$$$$$\mathrm{\Delta }\mathrm{z}\mathbf{OBC}$$$${\mathbf{BC}}^2={\mathbf{OC}}^2+{\mathbf{OB}}^2-2.\mathbf{OC}.\mathbf{OB}\cos (60-\mathbf{x})$$$$=10-6(\cos 60\cos \mathbf{x}+\sin 60\sin \mathbf{x})$$$$=10-3(\cos \mathbf{x}+\sqrt{3}\sin \mathbf{x})$$$$$$$$\mathbf{BC}=\sqrt{10-3(\cos \mathbf{x}+\sqrt{3}\sin \mathrm{x})}$$$$$$$${\mathbf{AC}}^2+{\mathbf{BC}}^2={\mathbf{AB}}^2$$$$19=26-10\cos \mathbf{x}+10-3\cos \mathbf{x}-3\sqrt{3}\sin \mathbf{x}$$$$13\cos \mathbf{x}+3\sqrt{3}\sin \mathbf{x}-36=-19$$$$$$$$13\cos \mathbf{x}+3\sqrt{3}\sin \mathbf{x}-17=0$$$$\mathbf{posonz}\mathbf{z}=\cos \mathbf{x}$$$$13\mathbf{z}+3\sqrt{3-3{\mathbf{z}}^2}-17=0$$$$\sqrt{3-{3\mathrm{z}}^2}={\left(\frac{17-13\mathrm{z}}{9}\right)}^2$$$$27-27{\mathbf{z}}^2={17}^2+169{\mathbf{z}}^2-442\mathbf{z}$$$$169{\mathbf{z}}^2-442\mathbf{z}+262=0$$$$$$$$\mathbf{z}=\cos \mathbf{x}=0,90\Rightarrow \mathbf{x}=24,77°$$$$$$$$\mathbf{Longeur}\mathbf{minimale}\mathbf{est}$$$$\mathbf{L}=\mathbf{AC}+\mathbf{BC}=$$$$\mathbf{AC}=\left(\sqrt{26-10\cos 24,77}\right)=4,113396$$$$$$$$\mathbf{BC}=\left(\sqrt{10-3(\cos 24,77+\sqrt{3}\sin 24,77}\right)$$$$=2,258083$$$$\mathbf{L}=4,113396+2,258083$$$$$$$$\mathbf{L}=6,371\mathbf{Km}$$$$$$$$$$

Commented by mr W last updated on 19/Dec/23

$$wrong!$$$$thereisnoreasonwhy\mathrm{\angle }ACB$$$$shouldbe90°.$$

Commented by Frix last updated on 19/Dec/23

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{we}\mathrm{can}\mathrm{get}\mathrm{an}\mathrm{exact}\mathrm{solution}.$$$$\mathrm{I}\mathrm{get}\approx 6.354\mathrm{km}\mathrm{with}\measuredangle COB=32.485°$$

Commented by mr W last updated on 19/Dec/23

$$agree!$$$$l_{min}\approx 6.35405km$$

Commented by mr W last updated on 19/Dec/23

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

$$\frac{\mathrm{d}(\mathrm{AB}+\mathrm{BC})}{\mathrm{dx}}=0$$$$\mathrm{d}\left(\sqrt{26-10\cos \mathbf{x}}\right)/\mathbf{dx}+$$$$\mathbf{d}\left(\sqrt{10-3(\cos \mathbf{x}+\sqrt{3}\sin \mathbf{x}}\right)/\mathbf{dx}=0$$$$\mathbf{soit}$$$$\frac{10\sin \mathbf{x}}{2\sqrt{36-10\cos \mathbf{x}}}+\frac{3(\sqrt{3}\cos \mathbf{x}-\sin \mathbf{x})}{2\sqrt{10-3(\cos \mathbf{x}+\sqrt{3}\sin \mathbf{x})}}=0$$$$\mathbf{resolutiin}\mathbf{de}\mathbf{l}\mathbf{equatiin}$$$$\mathbf{Soit}\mathbf{c}\mathbf{valeur}\mathbf{de}\mathbf{x}\mathbf{cherche}$$$$\{\begin{array}{l}\sin \mathbf{c}=\mathbf{h}\\ \cos \mathbf{c}=\mathbf{k}\end{array}$$$$\mathbf{Alors}$$$$(\mathbf{AB}+\mathbf{BC})=$$$$\sqrt{26-10\mathbf{k}}+\sqrt{10-3(\mathbf{k}+\sqrt{3}\mathbf{h}}$$$$(\mathit{I}\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}\mathit{t}\mathit{h}\mathbf{a}\mathit{t}\mathbf{this}\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{e}\mathit{c}\mathbf{a}\mathit{n}\mathit{r}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{p}\mathit{r}\mathit{o}\mathit{b}\mathit{l}\mathit{e}\mathit{m}\mathit{e})$$

Commented by mr W last updated on 19/Dec/23

$$yes,thisisarightapproach.$$

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