Question

In a triangle $\angle A={55}^0,\angle B={15}^0,\angle C={110}^{0}and\frac{c^2-a^2}{ab}=\lambda$ then $\lambda$

• A. 1
• B. 2
• C. 3
• D. none of these
  

Answer 1

The correct option is A 1

$\frac{c^2-a^2}{ab}=\lambda$

$\Rightarrow \frac{(c-a)(c+a)}{ab}=\lambda$

using Sine rule : $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\text{so}\frac{(k\sin C-K\sin A)(K\sin C+K\sin A)}{\left(K\sin A\right)\left(K\sin B\right)}=\lambda$

$\Rightarrow \frac{(\sin C-\sin A)(\sin C+\sin A)}{\sin A\cdot \sin B}=\lambda$

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

$\Rightarrow \frac{\left\{2\sin \left(\frac{c+A}{2}\right)\cos \left(\frac{A+c}{2}\right)\right\}\left\{2\sin \left(\frac{(-A)}{2}\right)\cos \left(\frac{c-A}{2}\right)\right\}}{\sin A\cdot \sin B}=\lambda$

$\Rightarrow \frac{\sin (A+C)\cdot \sin (C-A)}{\sin A\cdot \sin (\pi -(A+c\left)\right)}=\lambda$

$\Rightarrow \frac{\sin (A+C)\cdot \sin (C-A)}{\sin A\cdot \sin (A+C)}=\lambda$

$\Rightarrow \frac{\sin {55}^{\circ }}{\sin {55}^{\circ }}=\lambda =1$

Answer: option (A)