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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 ⇒ C=B−2A A+B+C=180° ⇒ C=180°−A−B B−2A=180°−A−B ⇒ A=2B−180° ⇒ C=360°−3B ⇒ 90°<B<120° ⇒ 0°<A<60° 0°<C<90° ⇒ B>A ∧ B>C let a=1 using only z^2 =x^2 +y^2 −2xycos(Z) leads to a=1 b=2 c=(3/2) so the triangle is unique and the rest is easy as you already have shown

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

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

(a/(SinA))=(b/(SinB))=(c/(SinC))=r b−c=(a/2) r(sinB−SinC)=((rSinA)/2) sinB−sinC=((sinA)/2) 2cos((B+C)/2).sin((B−C)/2)=sin(A/2)cos(A/2) 2cos((Π−A)/2).sin((B−C)/2)=sin(A/2).cos(A/2) 2sin(A/2)sin((B−C)/2)=sin(A/2)cos(A/2) sin((B−C)/2)=(1/2)cos(A/2) given ((B−C)/2)=A sin(((B−C)/2))=sinA=2sin(A/2)cos(A/2) (1/2)cos(A/2)=2sin(A/2)cos(A/2) sin(A/2)=(1/4) B+C=Π−A B−C=2A 2B=Π+A B=(Π/2)+(A/2) tanB=tan((Π/2)+(A/2)) tanB=−cot(A/2) sin(A/2)=(1/4) cot(A/2)=(((√(15)) )/1) tanB=−(√(15)) ((2tan(B/2))/(1−tan^2 (B/2)))=−(√(15)) ((2x)/(1−x^2 ))=−(√(15)) −(√(15)) +(√(15)) x^2 −2x=0 (√(15)) x^2 −2x−(√(15)) =0 x=((2±(√(4+60)) )/(2(√(15)) ))=((2±8)/(2(√(15))))=((10)/(2(√(15)))) and((−6)/(2(√(15)))) tan(B/2)=(5/(√(15))) and((−3)/(√(15))) angle B=(Π/2)+(A/2) so (B/2) lies in first quadrant so tan(B/2) can not be ((−3)/((√(15)) ))

$$\frac{{a}}{{SinA}}=\frac{{b}}{{SinB}}=\frac{{c}}{{SinC}}={r} \\ $$$${b}−{c}=\frac{{a}}{\mathrm{2}} \\ $$$${r}\left({sinB}−{SinC}\right)=\frac{{rSinA}}{\mathrm{2}} \\ $$$${sinB}−{sinC}=\frac{{sinA}}{\mathrm{2}} \\ $$$$\mathrm{2}{cos}\frac{{B}+{C}}{\mathrm{2}}.{sin}\frac{{B}−{C}}{\mathrm{2}}={sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}} \\ $$$$\mathrm{2}{cos}\frac{\Pi−{A}}{\mathrm{2}}.{sin}\frac{{B}−{C}}{\mathrm{2}}={sin}\frac{{A}}{\mathrm{2}}.{cos}\frac{{A}}{\mathrm{2}} \\ $$$$\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{sin}\frac{{B}−{C}}{\mathrm{2}}={sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}} \\ $$$${sin}\frac{{B}−{C}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${given}\:\:\frac{{B}−{C}}{\mathrm{2}}={A} \\ $$$${sin}\left(\frac{{B}−{C}}{\mathrm{2}}\right)={sinA}=\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}}=\mathrm{2}{sin}\frac{{A}}{\mathrm{2}}{cos}\frac{{A}}{\mathrm{2}} \\ $$$${sin}\frac{{A}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${B}+{C}=\Pi−{A} \\ $$$${B}−{C}=\mathrm{2}{A} \\ $$$$\mathrm{2}{B}=\Pi+{A} \\ $$$${B}=\frac{\Pi}{\mathrm{2}}+\frac{{A}}{\mathrm{2}}\:\: \\ $$$${tanB}={tan}\left(\frac{\Pi}{\mathrm{2}}+\frac{{A}}{\mathrm{2}}\right) \\ $$$${tanB}=−{cot}\frac{{A}}{\mathrm{2}} \\ $$$${sin}\frac{{A}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:{cot}\frac{{A}}{\mathrm{2}}=\frac{\sqrt{\mathrm{15}}\:}{\mathrm{1}} \\ $$$${tanB}=−\sqrt{\mathrm{15}}\: \\ $$$$\frac{\mathrm{2}{tan}\frac{{B}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{B}}{\mathrm{2}}}=−\sqrt{\mathrm{15}}\:\: \\ $$$$\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }=−\sqrt{\mathrm{15}}\: \\ $$$$−\sqrt{\mathrm{15}}\:+\sqrt{\mathrm{15}}\:{x}^{\mathrm{2}} −\mathrm{2}{x}=\mathrm{0} \\ $$$$\sqrt{\mathrm{15}}\:{x}^{\mathrm{2}} −\mathrm{2}{x}−\sqrt{\mathrm{15}}\:=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{60}}\:\:}{\mathrm{2}\sqrt{\mathrm{15}}\:}=\frac{\mathrm{2}\pm\mathrm{8}}{\mathrm{2}\sqrt{\mathrm{15}}}=\frac{\mathrm{10}}{\mathrm{2}\sqrt{\mathrm{15}}}\:{and}\frac{−\mathrm{6}}{\mathrm{2}\sqrt{\mathrm{15}}} \\ $$$${tan}\frac{{B}}{\mathrm{2}}=\frac{\mathrm{5}}{\sqrt{\mathrm{15}}}\:{and}\frac{−\mathrm{3}}{\sqrt{\mathrm{15}}}\:\: \\ $$$${angle}\:{B}=\frac{\Pi}{\mathrm{2}}+\frac{{A}}{\mathrm{2}}\:\:{so}\:\frac{{B}}{\mathrm{2}}\:\:{lies}\:{in}\:{first}\:{quadrant} \\ $$$${so}\:{tan}\frac{{B}}{\mathrm{2}}\:{can}\:{not}\:{be}\:\frac{−\mathrm{3}}{\sqrt{\mathrm{15}}\:} \\ $$$$ \\ $$

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

tg(B/2)=((√(15))/3)⇒B=52.24^• ×2=104.49 sin(A/2)=(1/4)⇒A=28.95^• C=B−2A=52.24×2−2×28.95=46.60 tg(B/2)=((√(15))/5)⇒B=37.76 C=37.76−2×28.95=−20.14 this is not feasible. but: tgB=−(√(15))⇒B=104.49^• C=B−2A=104.49−2×28.95=46.60^• tg(B/2)=tg((104.49)/2)=1.3=((√(15))/3). and this is acceptable. thank you for working dear tanmay.

$${tg}\frac{{B}}{\mathrm{2}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{3}}\Rightarrow{B}=\mathrm{52}.\mathrm{24}^{\bullet} ×\mathrm{2}=\mathrm{104}.\mathrm{49} \\ $$$${sin}\frac{{A}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{A}=\mathrm{28}.\mathrm{95}^{\bullet} \\ $$$${C}={B}−\mathrm{2}{A}=\mathrm{52}.\mathrm{24}×\mathrm{2}−\mathrm{2}×\mathrm{28}.\mathrm{95}=\mathrm{46}.\mathrm{60} \\ $$$${tg}\frac{{B}}{\mathrm{2}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{5}}\Rightarrow{B}=\mathrm{37}.\mathrm{76} \\ $$$${C}=\mathrm{37}.\mathrm{76}−\mathrm{2}×\mathrm{28}.\mathrm{95}=−\mathrm{20}.\mathrm{14} \\ $$$${this}\:{is}\:{not}\:{feasible}. \\ $$$${but}: \\ $$$${tgB}=−\sqrt{\mathrm{15}}\Rightarrow{B}=\mathrm{104}.\mathrm{49}^{\bullet} \\ $$$${C}={B}−\mathrm{2}{A}=\mathrm{104}.\mathrm{49}−\mathrm{2}×\mathrm{28}.\mathrm{95}=\mathrm{46}.\mathrm{60}^{\bullet} \\ $$$${tg}\frac{{B}}{\mathrm{2}}={tg}\frac{\mathrm{104}.\mathrm{49}}{\mathrm{2}}=\mathrm{1}.\mathrm{3}=\frac{\sqrt{\mathrm{15}}}{\mathrm{3}}. \\ $$$${and}\:{this}\:{is}\:{acceptable}. \\ $$$${thank}\:{you}\:{for}\:{working}\:{dear}\:{tanmay}. \\ $$

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

thank you dear tanmay. a=2R.sinA

$${thank}\:{you}\:{dear}\:{tanmay}. \\ $$$${a}=\mathrm{2}{R}.{sinA} \\ $$

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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