Question

In a triangle, the lengths of two larger sides are $10$ cm and $9$ cm. If the angles of the triangle are in $AP$, then the length of the third side is

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

Answer 1

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

Let in the figure, $\angle A>\angle B>\angle C$
Angle are in A.P.
Therefore, $\angle C=\theta -d,\angle B=\theta$
and $\angle A=\theta +d$
As $\angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow \theta +d+\theta +\theta -d={180}^{\circ }$
$\Rightarrow 3\theta ={180}^{\circ }$
$\Rightarrow \theta ={60}^{\circ }$
Now in $△ABC$, we have
$\cos B=\frac{a^2+c^2-b^2}{2ac}$
$\cos {60}^{\circ }=\frac{(10{)}^2+x^2-9^2}{2\times 10\times x}$
$\Rightarrow \frac{1}{2}=\frac{100+x^2-9^2}{20x}$
$\Rightarrow x^2-10x+19=0$
$\Rightarrow x=\frac{10\pm \sqrt{100-76}}{2}$
$=\frac{10\pm \sqrt{24}}{2}=5\pm \sqrt{6}$.