Question

In a triangle, the angle are in AP and the length of the two larger sides are $10$ and $9$ respectively, then the length of the third side can be

• A. $5\pm \sqrt{6}$
• B. $0.7$
• C. $1-\sqrt{6}$
• D. None of these

Answer 1

Answer: (A)

$5\pm \sqrt{6}$

Let the angles be $A=x-d,B=x,C=x+d$

Then, $x-d+x+x+d={180}^{\circ }$

$\Rightarrow 3x={180}^{\circ }\Rightarrow x={60}^{\circ }$

Therefore, two larger angles are $B$ and $C$.

Hence, $b=9$ and $c=10$.

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

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

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

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