If-in-a-ABC-a-2-b-2-a-2-b-2-sin-A-B-sin-A-B-then-the-triangle-is-

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

If in a △ A B C, \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{\sin (A - B)}{\sin (A + B)},

then the triangle is

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

Either isoscale or rt . angle

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

We have \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{\sin (A - B)}{\sin (A + B)}

\Leftrightarrow \frac{a^{2} - b^{2}}{\sin (A - B)} = \frac{a^{2} + b^{2}}{\sin (A + B)} = \frac{{2 a}^{2}}{\sin (A - B) + \sin (A + B)} = \frac{{2 b}^{2}}{\sin (A + B) - \sin (A - B)}

\Rightarrow \frac{{2 a}^{2}}{2 sinAcosB} = \frac{{2 b}^{2}}{2 cosAsinB} (1)

From sine theorem \frac{a}{sinA} = \frac{b}{sinB} we get

(1) \Leftrightarrow \frac{{4 R}^{2} \sin^{2} A}{sinAcosB} = \frac{{4 R}^{2} \sin^{2} B}{sinBcosA}

\Leftrightarrow \frac{sinA}{cosB} = \frac{sinB}{cosA} \Leftrightarrow 2 sinAcosA = 2 sinBcosB

\Leftrightarrow \sin 2 A = \sin 2 B \Leftrightarrow 2 A = 2 B (2) or

2 A = 180 ° - 2 B (3)

(2) \Leftrightarrow A = B \Leftrightarrow Δ ABC is isosceles at C

(3) \Leftrightarrow A = 90 ° - B \Leftrightarrow Δ ABC is right at C