Question

In a triangle, 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. $\pm \sqrt{6}$

Answer 1

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

A,B,C are in A.P.

Let the angles be $(a-d),a,(a+d)$

B= $\frac{A+C}{2}$

i.e. $\angle B$ =60$°$

Now, $CosB$ = $\frac{a^2+81-100}{2\times a\times 9}$

$a^2-19$ =18a

$a^2-9a-19$=0

a = $\frac{9\pm \sqrt{81-4\times (+9)}}{2\times 1}$

= $\frac{9\pm \sqrt{5}}{2}$

Now, let $b=9,c=10$

$\therefore$

$a^2+100-81-a$$\times$10

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

a= $\frac{10\pm \sqrt{100-19\times 4}}{2}$

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

Hence, $5\pm \sqrt{6}$ is the correct answer.