Question

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

• A. $5+\sqrt{6}$
• B. $0.7$
• C. $5-\sqrt{6}$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A $5+\sqrt{6}$
C $5-\sqrt{6}$

Let the angles of the $△$ in A.P be given by

$\angle A=\theta -\varphi ,B=\theta ,\angle C=\theta +\varphi$

$\therefore A+B+C={180}^0$

$\Rightarrow \theta ={60}^0=\angle B={60}^0$

Here two larger sides are $c=10$ and $b=9$

now $\cos B=\cos {60}^0=\left(\frac{a^2+b^2+c^2}{2ac}\right)$

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

$\therefore a=\frac{1}{2}[10\pm \sqrt{100-76}]=(5\pm \sqrt{6})$

Both these value of a are positive and smaller than b and c