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Question Number 184573 by mnjuly1970 last updated on 08/Jan/23

Commented by SEKRET last updated on 08/Jan/23

$$\mathbf{use}\mathbf{an}\mathbf{triangle}$$

Commented by Frix last updated on 08/Jan/23

$$a=0b=\pm 5c=\mp 8$$$$a=\pm \frac{40\sqrt{129}}{129}b=\pm \frac{25\sqrt{129}}{129}c=\pm \frac{64\sqrt{129}}{129}$$$${(a+b+c)}^2=9\vee 129$$

Answered by mr W last updated on 09/Jan/23

$$ihavefoundageneralsolutionfor$$$$thisequationsystem,seeQ164174.$$$$incurrentcaseweget$$$${(a+b+c)}^2=\frac{25+49+64+\sqrt{3(5+7+8)(-5+7+8)(5-7+8)(5+7-8)}}{2}$$$$=129$$

Answered by SEKRET last updated on 08/Jan/23

$$\{\begin{array}{l}(\mathbf{a}-\mathbf{b})({\mathbf{a}}^2+\mathbf{ab}+{\mathbf{b}}^2)=25(\mathbf{a}-\mathbf{b})\\ (\mathbf{b}-\mathbf{c})({\mathbf{b}}^2+\mathbf{bc}+{\mathbf{c}}^2)=49(\mathbf{b}-\mathbf{c})\\ (\mathbf{c}-\mathbf{a})({\mathbf{c}}^2+\mathbf{ac}+{\mathbf{a}}^2)=64(\mathbf{c}-\mathbf{a})\end{array}$$$$+\{\begin{array}{l}{\mathbf{a}}^3-{\mathbf{b}}^3=25\mathbf{a}-25\mathbf{b}\\ {\mathbf{b}}^3-{\mathbf{c}}^3=49\mathbf{b}-49\mathbf{c}\\ {\mathbf{c}}^3-{\mathbf{a}}^3=64\mathbf{c}-64\mathbf{a}\end{array}$$$$-39\mathbf{a}+24\mathbf{b}+15\mathbf{c}=0$$$$\mathbf{a}=\frac{5\mathbf{c}+8\mathbf{b}}{13}$$$$\{\begin{array}{l}{\left(\frac{5\mathbf{c}+8\mathbf{b}}{13}\right)}^2+\mathbf{b}\cdot \frac{5\mathbf{c}+8\mathbf{b}}{13}+{\mathbf{b}}^2=25\\ {\mathbf{b}}^2+\mathbf{bc}+{\mathbf{c}}^2=49\end{array}$$$$$$$$..........$$$$....$$

Answered by SEKRET last updated on 08/Jan/23

$$\mathbf{a}=0\mathbf{b}=-5\mathbf{c}=8$$$$\mathbf{a}=0\mathbf{b}=5\mathbf{c}=-8$$$$\mathbf{a}=\frac{40}{\sqrt{129}}\mathbf{b}=\frac{25}{\sqrt{129}}\mathbf{c}=\frac{64}{\sqrt{129}}$$$$\mathbf{a}=\frac{-40}{\sqrt{129}}\mathbf{b}=\frac{-25}{\sqrt{129}}\mathbf{c}=-\frac{64}{\sqrt{129}}$$

Commented by mr W last updated on 09/Jan/23

$$onlya,b,c>0areneeded.$$$$buthowdidyougetthis?$$

Answered by Rasheed.Sindhi last updated on 09/Jan/23

$$5^2=a^2+b^2-2ab\cos 120$$$$7^2=b^2+c^2-2bc\cos 120$$$$8^2=c^2+a^2-2ca\cos 120$$$$.....$$$$...$$

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