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Question Number 40948 by behi83417@gmail.com last updated on 30/Jul/18

Commented by MJS last updated on 30/Jul/18

$$B-C=2A\Rightarrow C=B-2A$$$$A+B+C=180°\Rightarrow C=180°-A-B$$$$B-2A=180°-A-B\Rightarrow A=2B-180°$$$$\Rightarrow C=360°-3B$$$$$$$$\Rightarrow 90°<B<120°$$$$\Rightarrow 0°<A<60°$$$$0°<C<90°$$$$\Rightarrow B>A\wedge B>C$$$$$$$$\mathrm{let}a=1$$$$\mathrm{using}\mathrm{only}{z}^2=x^2+y^2-2xy\cos \left(Z\right)\mathrm{leads}\mathrm{to}$$$$a=1$$$$b=2$$$$c=\frac{3}{2}$$$$\mathrm{so}\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{unique}\mathrm{and}\mathrm{the}\mathrm{rest}\mathrm{is}\mathrm{easy}$$$$\mathrm{as}\mathrm{you}\mathrm{already}\mathrm{have}\mathrm{shown}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18

$$\frac{a}{SinA}=\frac{b}{SinB}=\frac{c}{SinC}=r$$$$b-c=\frac{a}{2}$$$$r(sinB-SinC)=\frac{rSinA}{2}$$$$sinB-sinC=\frac{sinA}{2}$$$$2cos\frac{B+C}{2}.sin\frac{B-C}{2}=sin\frac{A}{2}cos\frac{A}{2}$$$$2cos\frac{\mathrm{\Pi }-A}{2}.sin\frac{B-C}{2}=sin\frac{A}{2}.cos\frac{A}{2}$$$$2sin\frac{A}{2}sin\frac{B-C}{2}=sin\frac{A}{2}cos\frac{A}{2}$$$$sin\frac{B-C}{2}=\frac{1}{2}cos\frac{A}{2}$$$$$$$$$$$$$$$$given\frac{B-C}{2}=A$$$$sin\left(\frac{B-C}{2}\right)=sinA=2sin\frac{A}{2}cos\frac{A}{2}$$$$\frac{1}{2}cos\frac{A}{2}=2sin\frac{A}{2}cos\frac{A}{2}$$$$sin\frac{A}{2}=\frac{1}{4}$$$$B+C=\mathrm{\Pi }-A$$$$B-C=2A$$$$2B=\mathrm{\Pi }+A$$$$B=\frac{\mathrm{\Pi }}{2}+\frac{A}{2}$$$$tanB=tan(\frac{\mathrm{\Pi }}{2}+\frac{A}{2})$$$$tanB=-cot\frac{A}{2}$$$$sin\frac{A}{2}=\frac{1}{4}cot\frac{A}{2}=\frac{\sqrt{15}}{1}$$$$tanB=-\sqrt{15}$$$$\frac{2tan\frac{B}{2}}{1-{tan}^2\frac{B}{2}}=-\sqrt{15}$$$$\frac{2x}{1-x^2}=-\sqrt{15}$$$$-\sqrt{15}+\sqrt{15}{x}^2-2x=0$$$$\sqrt{15}{x}^2-2x-\sqrt{15}=0$$$$x=\frac{2\pm \sqrt{4+60}}{2\sqrt{15}}=\frac{2\pm 8}{2\sqrt{15}}=\frac{10}{2\sqrt{15}}and\frac{-6}{2\sqrt{15}}$$$$tan\frac{B}{2}=\frac{5}{\sqrt{15}}and\frac{-3}{\sqrt{15}}$$$$angleB=\frac{\mathrm{\Pi }}{2}+\frac{A}{2}so\frac{B}{2}liesinfirstquadrant$$$$sotan\frac{B}{2}cannotbe\frac{-3}{\sqrt{15}}$$$$$$

Commented by behi83417@gmail.com last updated on 30/Jul/18

$$tg\frac{B}{2}=\frac{\sqrt{15}}{3}\Rightarrow B=52.{24}^{\bullet }\times 2=104.49$$$$sin\frac{A}{2}=\frac{1}{4}\Rightarrow A=28.{95}^{\bullet }$$$$C=B-2A=52.24\times 2-2\times 28.95=46.60$$$$tg\frac{B}{2}=\frac{\sqrt{15}}{5}\Rightarrow B=37.76$$$$C=37.76-2\times 28.95=-20.14$$$$thisisnotfeasible.$$$$but:$$$$tgB=-\sqrt{15}\Rightarrow B=104.{49}^{\bullet }$$$$C=B-2A=104.49-2\times 28.95=46.{60}^{\bullet }$$$$tg\frac{B}{2}=tg\frac{104.49}{2}=1.3=\frac{\sqrt{15}}{3}.$$$$andthisisacceptable.$$$$thankyouforworkingdeartanmay.$$

Commented by behi83417@gmail.com last updated on 30/Jul/18

$$thankyoudeartanmay.$$$$a=2R.sinA$$

Commented by tanmay.chaudhury50@gmail.com last updated on 30/Jul/18

Answered by behi83417@gmail.com last updated on 30/Jul/18

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