In any $\triangle ABC$,$a(b \cos C - c \cos B)$ equals

The angles of a triangle are in the ratio $5:1:6$. The ratio of the smallest side to the greatest side is:

Let $ABC$ be a triangle such that $\angle ACB = \frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value$(s)$ of $x$ for which $a = x^2+x+1, b = x^2-1$ and $c = 2x+1$ is (are)

The area of the triangle $ABC$ is $10\sqrt{3} \text{ cm}^2$,angle $B$ is $60^{\circ}$ and its perimeter is $20 \text{ cm}$. Then $\ell(AC) = $ (in $\text{ cm}$)

In a triangle $ABC$,if $r_1 r_2 + r_3 r = 35$,$r_2 r_3 + r r_1 = 63$,and $r_3 r_1 + r r_2 = 45$,then $2s =$

If the sides of a triangle are in the ratio $\sqrt{3} : \sqrt{5} : \sqrt{8+\sqrt{15}}$,then the largest angle in that triangle is