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 $^{\circ}$)

If $p_1, p_2, p_3$ are the altitudes and $a=4, b=5, c=6$ are the sides of a triangle $ABC$,then $\frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} =$

If in $\Delta ABC,$ $a = 6, b = 3$ and $\cos(A - B) = \frac{4}{5},$ then its area will be ..... $square \, unit.$

$AB$ is a vertical tower. The point $A$ is on the ground and $C$ is the middle point of $AB$. The part $CB$ subtends an angle $\alpha$ at a point $P$ on the ground. If $AP = n \cdot AB$,then the correct relation is:

If the angles $A, B$ and $C$ of a triangle are in an Arithmetic Progression and if $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively,then the value of the expression $\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$ is

If $ABCD$ is a cyclic quadrilateral,then the value of $\cos A - \cos B + \cos C - \cos D = $