Question

In a $△ABC$, the lengths of the two larger sides are $10$ and $9$ units, respectively. 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. None of these

Answer 1

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

Let $A,B$ and $C$ be the three angles of $△ABC$ and
Let $a=10$ and $b=9$
It is given that the angles are in AP.
Therefore, $2B=A+C$ on adding $B$ on both sides, we get
$3B=A+B+C$
$\Rightarrow 3B={180}^o\Rightarrow B={60}^o$
Now we know $\cos B=\frac{a^2+c^2-b^2}{2ac}$
$\Rightarrow \cos {60}^o=\frac{{10}^2+c^2-9^2}{2\times 10\times c}$

$\Rightarrow \frac{1}{2}=\frac{100+c^2-81}{20c}$

$\Rightarrow c^2-10c+19=0$

$\Rightarrow c=5\pm \sqrt{6}$