Question

In $\Delta ABC,\frac{b^2+c^2}{b^2-c^2}=\frac{\sin (B+C)}{\sin (B-C)}$ then the triangle is

• A. A right angled triangle
• B. An isosceles triangle
• C. A right angled isosceles triangle
• D. A scalene triangle

Answer 1

The correct option is A A right angled triangle

We have, $A+B+C=\pi$

$b=K\sin B$

$c=K\sin C$

Now,
$\frac{{\sin }^{2}B+{\sin }^{2}C}{{\sin }^{2}B-{\sin }^{2}C}=\frac{\sin A}{\sin (B-C)}$

$\frac{{\sin }^{2}B+{\sin }^{2}C}{(\sin B-\sin C)(\sin B+\sin C)}=\frac{\sin A}{\sin (B-C)}$

$\Rightarrow \frac{{\sin }^{2}B+{\sin }^{2}C}{2\sin \left(\frac{C+B}{2}\right)\cos \left(\frac{B-C}{2}\right)\times 2\cos \left(\frac{B+C}{2}\right)\times \sin \left(\frac{B-C}{2}\right)}=\frac{\sin A}{\sin (B-C)}$

$\Rightarrow \frac{{\sin }^{2}B+{\sin }^{2}C}{\sin (B-C)\sin (B+C)}=\frac{\sin A}{\sin (B-C)}$

$\Rightarrow {\sin }^{2}A={\sin }^{2}B+{\sin }^{2}C$

$\Rightarrow {\sin }^{2}C={\cos }^{2}B-{\cos }^{2}A$

$=(\cos B-\cos A)(\cos A+\cos B)$

$=2\times 2\cos \left(\frac{B+A}{2}\right)\cdot \cos \left(\frac{B-C}{2}\right)\cdot \sin \left(\frac{A+B}{2}\right)\cdot \sin \left(\frac{A-B}{2}\right)$

$\Rightarrow {\sin }^{2}C=\sin (A+B)\cdot \sin (A-B)$

$\Rightarrow {\sin }^{2}C=\sin C\cdot \sin (A-B)$

$\sin C=\sin (A-B)$

$C=A-B$

$A=C+B$

$A+B+C=\pi$

$\therefore A=\frac{\pi }{2}$