Question

In a $△ABC,\angle A={55}^o,\angle B={15}^o,\angle C={110}^{o}then{c}^2-a^2$ is equal to

• A. ab
• B. 2ab
• C. -ab
• D. None of these

Answer 1

The correct option is A

ab

$\frac{c}{sin{110}^o}=\frac{b}{sin{15}^o}=\frac{a}{sin{55}^o}=k$
Now $c^2-a^2=k^{2}si{n}^2{110}^o-k^{2}si{n}^2{55}^o$
= $k^2(sin{110}^o+sin{15}^o)(sin{110}^o-sin{15}^o)$
= $k^2\left(2sin\frac{{165}^o}{2}cos\frac{{55}^o}{2}\right)\left(2cos\frac{{165}^o}{2}sin\frac{{55}^o}{2}\right)$
= $k^{2}sin{165}^{o}sin{55}^o$
= $\left(ksin{15}^o\right)\left(ksin{55}^o\right)=ab$