Question

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

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

Answer 1

Answer: (A), (D)

$5-\sqrt{6}$
$5+\sqrt{6}$

Since the angles $A,B,C$ of $△ABC$ are in $A.P.,$thus $\angle B={60}^0$

$\cos B=\frac{a^2+c^2-b^2}{2ac}$

$\therefore \frac{1}{2}=\frac{{10}^2+c^2-9^2}{20c}$

$\Rightarrow c^2-10c+19=0\Rightarrow c=\frac{10\pm \sqrt{100-76}}{2}=5\pm \sqrt{6}$

so $\left(a\right)$ and $\left(d\right)$ are the correct solutions.