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

$$\mathrm{If}\mathrm{in}\mathrm{a}△ABC,\frac{a^2-b^2}{a^2+b^2}=\frac{\sin (A-B)}{\sin (A+B)},$$$$\mathrm{then}\mathrm{the}\mathrm{triangle}\mathrm{is}$$

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

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

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

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

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