Question

In a triangle $\angle A={55}^{\circ }$ and $\angle B={15}^{\circ }$, then $\frac{c^2-a^2}{ab}$ is equal to

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

Answer: (D)

1

Given $\angle A={55}^o,\angle B={15}^o$
$\Rightarrow \angle C={110}^0$ [sum of angles$={180}^o$]
Consider $\frac{c^2-a^2}{ab}$
$=\frac{4R^2({\sin }^{2}C-{\sin }^{2}A)}{4R^2\sin A\sin B}$ (by sine rule)
$=\frac{\sin (C+A)\sin (C-A)}{\sin A\sin B}$
$=\frac{\sin {165}^o\sin {55}^o}{\sin {55}^o\sin {15}^o}$
$=\frac{\sin ({180}^o-{165}^o)}{\sin {15}^o}$
$=\frac{\sin {15}^o}{\sin {15}^o}$
$=1$