Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 113991 by deepraj123 last updated on 16/Sep/20

If in a △ABC, ((a^2 −b^2 )/(a^2 +b^2 )) = ((sin (A−B))/(sin (A+B))) , then the triangle is

$$\mathrm{If}\:\:\mathrm{in}\:\mathrm{a}\:\bigtriangleup{ABC},\:\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{sin}\:\left({A}−{B}\right)}{\mathrm{sin}\:\left({A}+{B}\right)}\:, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is} \\ $$

Commented by som(math1967) last updated on 16/Sep/20

Either isoscale or rt.angle

$$\mathrm{Either}\:\mathrm{isoscale}\:\mathrm{or}\:\mathrm{rt}.\mathrm{angle} \\ $$

Answered by 1549442205PVT last updated on 17/Sep/20

We have ((a^2 −b^2 )/(a^2 +b^2 )) = ((sin (A−B))/(sin (A+B))) ⇔((a^2 −b^2 )/(sin(A−B)))=((a^2 +b^2 )/(sin(A+B)))=((2a^2 )/(sin(A−B)+sin(A+B)))=((2b^2 )/(sin(A+B)−sin(A−B))) ⇒((2a^2 )/(2sinAcosB))=((2b^2 )/(2cosAsinB))(1) From sine theorem (a/(sinA))=(b/(sinB)) we get (1)⇔((4R^2 sin^2 A)/(sinAcosB))=((4R^2 sin^2 B)/(sinBcosA)) ⇔((sinA)/(cosB))=((sinB)/(cosA))⇔2sinAcosA=2sinBcosB ⇔sin2A=sin2B⇔2A=2B (2) or 2A=180°−2B(3) (2)⇔A=B⇔ΔABC is isosceles at C (3)⇔A=90°−B⇔ΔABC is right at C

$$\mathrm{We}\:\mathrm{have}\:\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{sin}\:\left({A}−{B}\right)}{\mathrm{sin}\:\left({A}+{B}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{b}^{\mathrm{2}} }{\mathrm{sin}\left(\mathrm{A}−\mathrm{B}\right)}=\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\mathrm{2a}^{\mathrm{2}} }{\mathrm{sin}\left(\mathrm{A}−\mathrm{B}\right)+\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\mathrm{2b}^{\mathrm{2}} }{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)−\mathrm{sin}\left(\mathrm{A}−\mathrm{B}\right)} \\ $$$$\Rightarrow\frac{\mathrm{2a}^{\mathrm{2}} }{\mathrm{2sinAcosB}}=\frac{\mathrm{2b}^{\mathrm{2}} }{\mathrm{2cosAsinB}}\left(\mathrm{1}\right) \\ $$$$\mathrm{From}\:\mathrm{sine}\:\mathrm{theorem}\:\frac{\mathrm{a}}{\mathrm{sinA}}=\frac{\mathrm{b}}{\mathrm{sinB}}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\frac{\mathrm{4R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{A}}{\mathrm{sinAcosB}}=\frac{\mathrm{4R}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{B}}{\mathrm{sinBcosA}} \\ $$$$\Leftrightarrow\frac{\mathrm{sinA}}{\mathrm{cosB}}=\frac{\mathrm{sinB}}{\mathrm{cosA}}\Leftrightarrow\mathrm{2sinAcosA}=\mathrm{2sinBcosB} \\ $$$$\Leftrightarrow\mathrm{sin2A}=\mathrm{sin2B}\Leftrightarrow\mathrm{2A}=\mathrm{2B}\:\:\left(\mathrm{2}\right)\:\mathrm{or} \\ $$$$\mathrm{2A}=\mathrm{180}°−\mathrm{2B}\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\Leftrightarrow\mathrm{A}=\mathrm{B}\Leftrightarrow\Delta\mathrm{ABC}\:\mathrm{is}\:\mathrm{isosceles}\:\mathrm{at}\:\mathrm{C} \\ $$$$\left(\mathrm{3}\right)\Leftrightarrow\mathrm{A}=\mathrm{90}°−\mathrm{B}\Leftrightarrow\Delta\mathrm{ABC}\:\mathrm{is}\:\mathrm{right}\:\mathrm{at}\:\mathrm{C} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com