Let $a, b, c$ be the lengths of sides of triangle $ABC$ such that $\frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k$. Then $\frac{(A(\triangle ABC))^2}{k^4}=$

If $\Delta$ denotes the area of triangle $ABC$,then $(b \sin C + c \sin B)(b \cos C + c \cos B) =$

In $\triangle ABC$,if $\sin^2 A + \sin^2 B = \sin^2 C$ and $l(AB) = 10$,then the maximum value of the area of $\triangle ABC$ is

Let $S = \left\{ \theta \in [-\pi, \pi] - \left\{ \pm \frac{\pi}{2} \right\} : \sin \theta \tan \theta + \tan \theta = \sin 2 \theta \right\}$. If $T = \sum_{\theta \in S} \cos 2 \theta$,then $T + n(S)$ is equal to:

$A$ tower $AB$ leans towards west making an angle $\alpha$ with the vertical. The angular elevation of $B$,the top most point of the tower,is $\beta$ as observed from a point $C$ due east of $A$ at a distance $d$ from $A$. If the angular elevation of $B$ from a point $D$ due east of $C$ at a distance $2d$ from $C$ is $\gamma$,then $2\tan \alpha$ can be given as

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 $^{\circ}$)