Question Number 54602 by behi83417@gmail.com last updated on 07/Feb/19
$${in}\:{a}\:{given}\:{triangle}: \\ $$$$\:\:\:\boldsymbol{\mathrm{tg}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}=\frac{\boldsymbol{\mathrm{a}}.\boldsymbol{\mathrm{tgA}}+\boldsymbol{\mathrm{b}}.\boldsymbol{\mathrm{tgB}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}\:. \\ $$$$\boldsymbol{\mathrm{define}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{kind}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}. \\ $$
Commented by MJS last updated on 08/Feb/19
$$\mathrm{using}\:\mathrm{the}\:\mathrm{laws}\:\mathrm{of}\:\mathrm{sines}\:\mathrm{and}\:\mathrm{cosines}\:\mathrm{I}\:\mathrm{get} \\ $$$${c}=\sqrt{{ab}+\sqrt{{a}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} +{b}^{\mathrm{4}} }} \\ $$$$\mathrm{but}\:\mathrm{this}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a}\:\mathrm{special}\:\mathrm{kind}\:\mathrm{of} \\ $$$$\mathrm{triangle}… \\ $$