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Question Number 203055 by mr W last updated on 08/Jan/24

Commented by mr W last updated on 08/Jan/24

$$isitpossibletodeterminethe$$$$unknownareaintermsofother$$$$partialareasA,B,C,D,E?$$

Commented by ajfour last updated on 09/Jan/24

$$ithink,yes.$$$$Weassumefouranglesandoneside$$$$asvariables.$$$$Systemisunique.Fiveequations$$$$canbewrittenusingthesegiven$$$$areastofindthem.$$

Commented by mr W last updated on 09/Jan/24

$$buttheequationsarenon-linear,$$$$seebelow.soi{don}^{\prime }tthinktheyare$$$$solvable.$$

Commented by mr W last updated on 12/Jan/24

$$fouranglesandoneside{don}^{\prime }tmake$$$$thesystemuniqueasfollowing$$$$figureshows.thoughthebluestar$$$$andthegreenstarhaveequalangles$$$$andonecommonequalside\left(red\right),$$$$buttheyaredifferent.$$

Commented by mr W last updated on 12/Jan/24

Answered by mr W last updated on 09/Jan/24

Commented by mr W last updated on 09/Jan/24

$$atfirst{let}^{\prime }sfindthesidelengthes$$$$ofatriangleifitsareaisSandits$$$$interioranglesare\alpha ,\beta and\gamma$$$$respectivelywith\alpha +\beta +\gamma =180°.$$$$a=2R\sin \alpha$$$$b=2R\sin \beta$$$$c=2R\sin \gamma$$$$S=\frac{abc}{4R}=\frac{8R^3\sin \alpha \sin \beta \sin \gamma }{4R}=2R^2\sin \alpha \sin \beta \sin \gamma$$$$\Rightarrow R=\sqrt{\frac{S}{2\sin \alpha \sin \beta \sin \gamma }}$$$$\Rightarrow a=\sqrt{\frac{2S\sin \alpha }{\sin \beta \sin \gamma }}$$$$\Rightarrow b=\sqrt{\frac{2S\sin \beta }{\sin \gamma \sin \alpha }}$$$$\Rightarrow c=\sqrt{\frac{2S\sin \gamma }{\sin \alpha \sin \beta }}$$

Commented by mr W last updated on 10/Jan/24

Commented by mr W last updated on 12/Jan/24

$$saytheanglesofthe5-pointstar$$$$are{\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4,{\theta }_{5}respectively.$$$${\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4+{\theta }_5=180°...\left(vi\right)$$$$soinfactwehaveonlyfour$$$$unknownangles.$$$${\phi }_1={\theta }_2+{\theta }_4$$$${\phi }_2={\theta }_3+{\theta }_5$$$${\phi }_3={\theta }_4+{\theta }_1$$$${\phi }_4={\theta }_5+{\theta }_2$$$${\phi }_5={\theta }_1+{\theta }_3$$$$$$$$saytheunknownareainthemiddle$$$$isX.$$$$accordingtoabovewehave$$$$a_1=\sqrt{\frac{2S_1\sin {\phi }_2}{\sin {\theta }_1\sin {\phi }_1}}=\sqrt{\frac{2S_1\sin ({\theta }_3+{\theta }_5)}{\sin {\theta }_1\sin ({\theta }_2+{\theta }_4)}}$$$$b_1=\sqrt{\frac{2S_1\sin {\phi }_1}{\sin {\theta }_1\sin {\phi }_2}}=\sqrt{\frac{2S_1\sin ({\theta }_2+{\theta }_4)}{\sin {\theta }_1\sin ({\theta }_3+{\theta }_5)}}$$$$c_1=\sqrt{\frac{2S_1\sin {\theta }_1}{\sin {\phi }_1\sin {\phi }_2}}=\sqrt{\frac{2S_1\sin {\theta }_1}{\sin ({\theta }_2+{\theta }_4)\sin ({\theta }_3+{\theta }_5)}}$$$$P_1{Q}_3=b_1+c_2=\sqrt{\frac{2S_1\sin ({\theta }_2+{\theta }_4)}{\sin {\theta }_1\sin ({\theta }_3+{\theta }_5)}}+\sqrt{\frac{2S_2\sin {\theta }_2}{\sin ({\theta }_3+{\theta }_5)\sin ({\theta }_4+{\theta }_1)}}$$$$P_1{Q}_3=\sqrt{\frac{2(S_1+S_4+X)\sin {\theta }_4}{\sin {\theta }_1\sin ({\theta }_1+{\theta }_4)}}$$$$\sqrt{\frac{S_1\sin ({\theta }_2+{\theta }_4)}{\sin {\theta }_1\sin ({\theta }_3+{\theta }_5)}}+\sqrt{\frac{S_2\sin {\theta }_2}{\sin ({\theta }_3+{\theta }_5)\sin ({\theta }_4+{\theta }_1)}}=\sqrt{\frac{(S_1+S_4+X)\sin {\theta }_4}{\sin {\theta }_1\sin ({\theta }_1+{\theta }_4)}}$$$$\Rightarrow \sqrt{S_1\sin ({\theta }_4+{\theta }_1)\sin ({\theta }_2+{\theta }_4)}+\sqrt{S_2\sin {\theta }_1\sin {\theta }_2}=\sqrt{(S_1+S_4+X)\sin {\theta }_4\sin ({\theta }_3+{\theta }_5)}...\left(i\right)$$$$similarly$$$$\sqrt{\frac{S_2\sin ({\theta }_3+{\theta }_5)}{\sin {\theta }_2\sin ({\theta }_4+{\theta }_1)}}+\sqrt{\frac{S_3\sin {\theta }_3}{\sin ({\theta }_4+{\theta }_1)\sin ({\theta }_5+{\theta }_2)}}=\sqrt{\frac{(S_2+S_5+X)\sin {\theta }_5}{\sin {\theta }_2\sin ({\theta }_2+{\theta }_5)}}$$$$\Rightarrow \sqrt{S_2\sin ({\theta }_5+{\theta }_2)\sin ({\theta }_3+{\theta }_5)}+\sqrt{S_3\sin {\theta }_2\sin {\theta }_3}=\sqrt{(S_2+S_5+X)\sin {\theta }_5\sin ({\theta }_4+{\theta }_1)}...\left(ii\right)$$$$\sqrt{\frac{S_3\sin ({\theta }_4+{\theta }_1)}{\sin {\theta }_3\sin ({\theta }_5+{\theta }_2)}}+\sqrt{\frac{S_4\sin {\theta }_4}{\sin ({\theta }_5+{\theta }_2)\sin ({\theta }_1+{\theta }_3)}}=\sqrt{\frac{(S_3+S_1+X)\sin {\theta }_1}{\sin {\theta }_3\sin ({\theta }_3+{\theta }_1)}}$$$$\Rightarrow \sqrt{S_3\sin ({\theta }_1+{\theta }_3)\sin ({\theta }_4+{\theta }_1)}+\sqrt{S_4\sin {\theta }_3\sin {\theta }_4}=\sqrt{(S_3+S_1+X)\sin {\theta }_1\sin ({\theta }_5+{\theta }_2)}...\left(iii\right)$$$$\sqrt{\frac{S_4\sin ({\theta }_5+{\theta }_2)}{\sin {\theta }_4\sin ({\theta }_1+{\theta }_3)}}+\sqrt{\frac{S_5\sin {\theta }_5}{\sin ({\theta }_1+{\theta }_3)\sin ({\theta }_2+{\theta }_4)}}=\sqrt{\frac{(S_4+S_2+X)\sin {\theta }_2}{\sin {\theta }_4\sin ({\theta }_4+{\theta }_2)}}$$$$\Rightarrow \sqrt{S_4\sin ({\theta }_2+{\theta }_4)\sin ({\theta }_5+{\theta }_2)}+\sqrt{S_5\sin {\theta }_4\sin {\theta }_5}=\sqrt{(S_4+S_2+X)\sin {\theta }_2\sin ({\theta }_1+{\theta }_3)}...\left(iv\right)$$$$\sqrt{\frac{S_5\sin ({\theta }_1+{\theta }_3)}{\sin {\theta }_5\sin ({\theta }_2+{\theta }_4)}}+\sqrt{\frac{S_1\sin {\theta }_1}{\sin ({\theta }_2+{\theta }_4)\sin ({\theta }_3+{\theta }_5)}}=\sqrt{\frac{(S_5+S_3+X)\sin {\theta }_3}{\sin {\theta }_5\sin ({\theta }_5+{\theta }_3)}}$$$$\Rightarrow \sqrt{S_5\sin ({\theta }_3+{\theta }_5)\sin ({\theta }_1+{\theta }_3)}+\sqrt{S_1\sin {\theta }_5\sin {\theta }_1}=\sqrt{(S_5+S_3+X)\sin {\theta }_3\sin ({\theta }_2+{\theta }_3)}...\left(v\right)$$$$$$$$wehave6equationsfor6unknowns:$$$${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4,{\theta }_{5}andX.$$

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