Question

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

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is D $1$

We know that $\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle C={180}^{\circ }-\angle A-\angle B$

$\Rightarrow \angle C={180}^{\circ }-{55}^{\circ }-{15}^{\circ }={110}^{\circ }$

Using Sine rule,$c=2R\sin C=2R\sin {110}^{\circ }$

$b=2R\sin B=2R\sin {15}^{\circ }$

$a=2R\sin A=2R\sin {55}^{\circ }$

Substituting the above in $\frac{c^2-a^2}{ab}$

$=\frac{4R^2{\sin }^2{110}^{\circ }-4R^2{\sin }^2{55}^{\circ }}{4R^2\sin {55}^{\circ }\sin {15}^{\circ }}$

$=\frac{{\sin }^2{110}^{\circ }-{\sin }^2{55}^{\circ }}{\sin {55}^{\circ }\sin {15}^{\circ }}$

$=\frac{\sin {165}^{\circ }\sin {55}^{\circ }}{\sin {55}^{\circ }\sin {15}^{\circ }}$

$=\frac{\sin ({180}^{\circ }-{15}^{\circ })\sin {55}^{\circ }}{\sin {55}^{\circ }\sin {15}^{\circ }}$

$=\frac{\sin {15}^{\circ }\sin {55}^{\circ }}{\sin {55}^{\circ }\sin {15}^{\circ }}$

$=\frac{\sin {15}^{\circ }}{\sin {15}^{\circ }}$

$=1$