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

In a triangle,if $r_1 = 2r_2 = 3r_3$,then $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}$ is equal to

In $\Delta ABC$,$a \cot A + b \cot B + c \cot C = . . . $ (where $r$ is inradius and $R$ is circumradius.)

In an isosceles triangle $ABC$,$\angle C = \angle A$. If the point of intersection of the bisectors of internal angles $\angle A$ and $\angle C$ divides the median of side $AC$ in the ratio $3 : 1$ (from vertex $B$ to side $AC$),then the value of $\csc \frac{B}{2}$ is equal to

In $\triangle ABC$,if $\angle A=90^{\circ}$,then $(r_2-r_1)(r_3-r_1)=$

$A$ tower is situated on a horizontal plane. Two points lie on the line passing through the base of the tower,at distances $a$ and $b$ from the base. The angles of elevation of the top of the tower from these points are $\alpha$ and $90^\circ - \alpha$. If the line segment joining the two points subtends an angle $\theta$ at the top of the tower,find the height of the tower.