In a triangle $ABC$,if $c^2-a^2=b(\sqrt{3}c-b)$ and $b^2-a^2=c(c-a)$,then $\angle ACB=$ (in $^{\circ}$)

In triangle $ABC$,$\frac{\tan A}{2} = \frac{\tan B}{3} = \frac{\tan C}{4}$,then the value of $\sec^2 A + \sec^2 B + \sec^2 C$ is:

The angle bisectors $BD$ and $CE$ of a $\triangle ABC$ are divided by the incentre $I$ in the ratios $3:2$ and $2:1$ respectively. Then,the ratio in which $I$ divides the angle bisector through $A$ is

In a $\Delta ABC$,the sides $a$,$b$,$c$ are the roots of the equation $x^3 - 11x^2 + 38x - 40 = 0$; then $\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = $

If angles $A$,$B$,and $C$ of a $\Delta ABC$ are $75^o$,$45^o$,and $60^o$ respectively,then the ratio of the areas of $\Delta OBC$,$\Delta COA$,and $\Delta AOB$ respectively is [where $O$ is the circumcentre of the triangle].

If the radius of the incircle of a triangle with sides $5k, 6k$ and $5k$ is $6$,then the largest angle of that triangle is