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Question Number 18904 by Satyamtt last updated on 01/Aug/17
Solve the triangle in which a=((√3)+1),   b=((√3)−1) and ∠C=60°.
$${Solve}\:{the}\:{triangle}\:{in}\:{which}\:{a}=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right),\: \\ $$$${b}=\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\:{and}\:\angle{C}=\mathrm{60}°. \\ $$$$ \\ $$
Answered by behi.8.3.4.1.7@gmail.com last updated on 01/Aug/17
c^2 =a^2 +b^2 −2ab.cosC=      =((√3)+1)^2 +((√3)−1)^2 −2.((√3)−1)((√3)+1).(1/2)=  =8−2=6⇒   c=(√6)  (a/(sinA))=(c/(sinC))⇒sinA=(((√3)+1)/( (√6))).((√3)/2)=((3+(√3))/(2(√6)))=  =((3(√6)+3(√2))/(12))=(((√6)+(√2))/4)⇒A=75^•   B=180−(60+75)=45^•  ■
$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}.{cosC}= \\ $$$$\:\:\:\:=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}.\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}+\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{2}}= \\ $$$$=\mathrm{8}−\mathrm{2}=\mathrm{6}\Rightarrow\:\:\:{c}=\sqrt{\mathrm{6}} \\ $$$$\frac{{a}}{{sinA}}=\frac{{c}}{{sinC}}\Rightarrow{sinA}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{6}}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{6}}}= \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{6}}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{12}}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\Rightarrow{A}=\mathrm{75}^{\bullet} \\ $$$${B}=\mathrm{180}−\left(\mathrm{60}+\mathrm{75}\right)=\mathrm{45}^{\bullet} \:\blacksquare \\ $$

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