Question

In a triangle, the lengths of two larger sides are $10$ and $9$.If the angles are in A.P, then the length of the third side can be

• A. $5\pm \sqrt{6}$
• B. $3\sqrt{3}$
• C. $5$
• D. $\sqrt{5}\pm 6$

Answer 1

The correct option is A $5\pm \sqrt{6}$

Let $a>b>c$
$\therefore a=10$ and $b=9$
According to the question,
$2B=A+C$
or $2B={180}^0-B$
$\therefore B={60}^0$
$\therefore \cos B=\frac{a^2+c^2-b^2}{2ac}$
$\Rightarrow \frac{1}{2}=\frac{100+c^2-81}{20c}$
$\Rightarrow 10c=c^2+19$
$\Rightarrow c^2-10c+19$ is quadratic in $c$
$\Rightarrow c=\frac{10\pm \sqrt{100-76}}{2}$
$=\frac{10\pm \sqrt{24}}{2}$
$\therefore c=5\pm \sqrt{6}$