The angles $\alpha, \beta, \gamma$ of a triangle satisfy the equations $2 \sin \alpha + 3 \cos \beta = 3 \sqrt{2}$ and $3 \sin \beta + 2 \cos \alpha = 1$. Then,angle $\gamma$ equals (in $^{\circ}$)

In a $\Delta ABC,$ let $\angle C = \frac{\pi}{2}.$ If $r$ and $R$ are the inradius and the circumradius respectively of the triangle,then $2(r + R)$ is equal to

If the median of a triangle $ABC$ through $A$ is perpendicular to $AB$,then the value of $\frac{\tan A}{\tan B}$ is equal to

In $\Delta ABC$,${a^2}({\cos ^2}B - {\cos ^2}C) + {b^2}({\cos ^2}C - {\cos ^2}A) + {c^2}({\cos ^2}A - {\cos ^2}B) = $

In a triangle $ABC$,$\angle B < \angle C$ and the values of $B$ and $C$ satisfy the equation $2 \tan x - k (1 + \tan^2 x) = 0$,where $0 < k < 1$. Then the measure of angle $A$ is:

Observe the following statements:
$(I)$ In $\triangle ABC$,$b \cos^2 \frac{C}{2} + c \cos^2 \frac{B}{2} = s$
$(II)$ In $\triangle ABC$,$\cot \frac{A}{2} = \frac{b+c}{a} \implies B = 90^{\circ}$
Which of the following is correct?