Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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^•  ■

$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}.{cosC}= \\ $$$$\:\:\:\:=\left(\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}.\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}+\mathrm{1}\right).\frac{\mathrm{1}}{\mathrm{2}}= \\ $$$$=\mathrm{8}−\mathrm{2}=\mathrm{6}\Rightarrow\:\:\:{c}=\sqrt{\mathrm{6}} \\ $$$$\frac{{a}}{{sinA}}=\frac{{c}}{{sinC}}\Rightarrow{sinA}=\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\sqrt{\mathrm{6}}}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{2}\sqrt{\mathrm{6}}}= \\ $$$$=\frac{\mathrm{3}\sqrt{\mathrm{6}}+\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{12}}=\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}\Rightarrow{A}=\mathrm{75}^{\bullet} \\ $$$${B}=\mathrm{180}−\left(\mathrm{60}+\mathrm{75}\right)=\mathrm{45}^{\bullet} \:\blacksquare \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com