Let-us-call-complex-triangle-which-has-either-sides-or-angles-are-complex-numbers-Let-a-b-c-R-which-are-sides-of-a-complex-triangle-which-need-not-satisfy-triangle-inequality-say-a-1-b-2-and-c-4-P

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Question Number 15759 by prakash jain last updated on 13/Jun/17

$$\mathrm{Let}\mathrm{us}\mathrm{call}\mathrm{complex}\mathrm{triangle}\mathrm{which}$$$$\mathrm{has}\mathrm{either}\mathrm{sides}\mathrm{or}\mathrm{angles}\mathrm{are}$$$$\mathrm{complex}\mathrm{numbers}.$$$$\mathrm{Let}a,b,c\in \mathbb{R}\mathrm{which}\mathrm{are}\mathrm{sides}\mathrm{of}$$$$\mathrm{a}\mathrm{complex}\mathrm{triangle}\mathrm{which}\mathrm{need}$$$$\mathrm{not}\mathrm{satisfy}\mathrm{triangle}\mathrm{inequality}.$$$$\mathrm{say}a=1,b=2\mathrm{and}c=4.$$$$\mathrm{Prove}\left(\mathrm{or}\mathrm{counter}\mathrm{example}\right)$$$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$$$$\mathrm{Is}A+B+C=\pi ?$$$$\mathrm{Assume}\mathrm{only}\mathrm{principle}\mathrm{solution}$$$$\mathrm{for}A,B\mathrm{and}C.$$$$\mathrm{For}\mathrm{such}\mathrm{a}\mathrm{triangle}A,B\mathrm{and}C$$$$\mathrm{will}\mathrm{take}\mathrm{complex}\mathrm{values}.$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Jun/17

$$a^2=b^2+c^2-2bccosA\Rightarrow$$$$cosA=\frac{4+16-1}{2\times 2\times 4}=\frac{19}{16}>1$$$$\Rightarrow \frac{e^{iA}+e^{-iA}}{2}=\frac{19}{16}\stackrel{e^{iA}=t}{\Rightarrow }16(t+\frac{1}{t})=38$$$$\Rightarrow 16t^2-38t+16=0\Rightarrow t=\frac{38\pm \sqrt{{38}^2-4\times 256}}{32}=1.83,.55$$$$\{\begin{array}{l}{e}^{iA}=1.82\Rightarrow iA=ln1.82\Rightarrow A=-0.6\mathit{i}\\ {\mathit{e}}^{\mathit{i}\mathit{A}}=.55\Rightarrow \mathit{i}\mathit{A}=ln.55\Rightarrow \mathit{A}=+0.6\mathit{i}\end{array}$$$$\mathit{c}\mathit{o}\mathit{s}\mathit{B}=\frac{a^2+c^2-b^2}{2ac}=\frac{1+16-4}{2\times 1\times 4}=\frac{13}{8}$$$$\frac{e^{iB}+e^{-iB}}{2}=\frac{13}{8}\stackrel{e^{iB}=s}{\Rightarrow }8(s+\frac{1}{s})=13$$$$\Rightarrow 8s^2-13s+8=0\Rightarrow s=\frac{13\pm \sqrt{169-4\times 64}}{16}$$$$\Rightarrow s=.81\pm .58\mathit{i}=\sqrt{.{81}^2+.{58}^2}\left(\frac{.81\pm .58i}{\sqrt{.{81}^2+.{58}^2}}\right)=$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\Rightarrow B=\{\begin{array}{l}{cos}^{-1}(.81)=35.{90}^{\bullet }\\ -{cos}^{-1}(.81)=-35.{90}^{\bullet }\end{array}$$$$cosC=\frac{a^2+b^2-c^2}{2ab}=\frac{1+4-16}{2\times 1\times 2}=-\frac{11}{4}$$$$\Rightarrow \frac{e^{iC}+e^{-iC}}{2}=-\frac{11}{4}\stackrel{e^{iC}=p}{\Rightarrow }\Rightarrow 4(p+\frac{1}{p})=-11$$$$\Rightarrow 4p^2+11p+4=0\Rightarrow p=\frac{-11\pm \sqrt{121-64}}{8}=$$$$\Rightarrow p=-.86,-4.6$$$$\Rightarrow e^{iC}=-.86=.86i^2\Rightarrow iC=ln.86+2lni=$$$$=-.15+i\pi \Rightarrow C=+.15i+\pi$$$$e^{iC}=-4.6=4.6i^2\Rightarrow iC=ln4.6+2lni=$$$$=1.52+i\pi \Rightarrow C=-1.52i+\pi$$$$\Rightarrow C=\{\begin{array}{l}.15\mathit{i}+\mathit{\pi }\\ -1.52\mathit{i}+\mathit{\pi }\end{array}$$$$(i=0+1\times i=cos90+isin90=e^{i\frac{\pi }{2}}\Rightarrow lni=i\frac{\pi }{2})$$$$\Rightarrow A=\{\begin{array}{l}-.6\mathit{i}\\ +.6\mathit{i}\end{array},B=\{\begin{array}{l}35.{90}^{\bullet }\\ -35.{90}^{\bullet }\end{array},C=\{\begin{array}{l}.15\mathit{i}+180\\ -1.52+180\end{array}$$$$\Rightarrow A+B+C=$$$$=-.6i+35.90+.15i+180=-.45\mathit{i}+215.90$$$$=.6\mathit{i}-35.90-1.52\mathit{i}+180=-.88\mathit{i}+144.1$$$$=-.6\mathit{i}+35.90-1.52\mathit{i}+180=-2.12\mathit{i}+215.90$$$$=-.6\mathit{i}-35.90+.15\mathit{i}+180=-.45\mathit{i}+144.1$$$$=-.6\mathit{i}-35.90-1.52+180=-2.12\mathit{i}+144.1$$$$=\dots \dots \dots \dots \dots .$$$$\mathit{t}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathit{i}\mathit{s}\mathit{n}\mathit{o}\mathit{c}\mathit{a}\mathit{s}\mathit{e}\mathit{f}\mathit{o}\mathit{r}:A+B+C=180.$$

Commented by prakash jain last updated on 14/Jun/17

$${\cos }^{-1}(-\frac{11}{4})+{\cos }^{-1}\left(\frac{19}{16}\right)+{\cos }^{-1}\left(\frac{13}{8}\right)=\pi$$$$\mathrm{I}\mathrm{think}\mathrm{you}\mathrm{mixed}\mathrm{degrees}\mathrm{and}\mathrm{radians}$$$$\mathrm{somewhere}.$$$$\mathrm{I}\mathrm{think}\mathrm{theory}\mathrm{of}\mathrm{triangle}\mathrm{remains}$$$$\mathrm{valid}\mathrm{even}\mathrm{with}\mathrm{complex}\mathrm{sides}$$$$\mathrm{and}\mathrm{angles}.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 15/Jun/17

$$mrprakash!{cos}^{-1}\left(\frac{19}{16}\right),notavalid$$$$anglewithprimaryelementsof$$$$geometryandtrigonometry.$$

Commented by prakash jain last updated on 18/Jun/17

$$\mathrm{Yes}.\mathrm{I}\mathrm{started}\mathrm{this}\mathrm{question}\mathrm{by}$$$$\mathrm{calling}\mathrm{these}\mathrm{triangle}\mathrm{complex}$$$$\mathrm{triangles}.$$

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