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Question Number 54602 by behi83417@gmail.com last updated on 07/Feb/19

$$inagiventriangle:$$$$\mathbf{tg}\frac{\mathbf{C}}{2}=\frac{\mathbf{a}.\mathbf{tgA}+\mathbf{b}.\mathbf{tgB}}{\mathbf{a}+\mathbf{b}}.$$$$\mathbf{define}\mathbf{the}\mathbf{kind}\mathbf{of}\mathbf{triangle}.$$

Commented by MJS last updated on 08/Feb/19

$$\mathrm{using}\mathrm{the}\mathrm{laws}\mathrm{of}\mathrm{sines}\mathrm{and}\mathrm{cosines}\mathrm{I}\mathrm{get}$$$$c=\sqrt{ab+\sqrt{a^4-a^2{b}^2+b^4}}$$$$\mathrm{but}\mathrm{this}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{seem}\mathrm{to}\mathrm{be}\mathrm{a}\mathrm{special}\mathrm{kind}\mathrm{of}$$$$\mathrm{triangle}...$$

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