Two vertical poles of equal heights are $120 \, m$ apart. On the line joining their bottoms,$A$ and $B$ are two points. The angle of elevation of the top of one pole from $A$ is $45^\circ$ and that of the other pole from $B$ is also $45^\circ$. If $AB = 30 \, m$,then the height of each pole is.....$m$.

$AB$ is a vertical pole with $B$ at the ground level and $A$ at the top. $A$ man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is $60^{\circ}$. He moves away from the pole along the line $BC$ to a point $D$ such that $CD = 7 \ m$. From $D$,the angle of elevation of the point $A$ is $45^{\circ}$. Then the height of the pole is

$A$ pole stands vertically inside a triangular park $ABC$. If the angle of elevation of the top of the pole from each corner of the park is the same,then in the $\Delta ABC$,the foot of the pole is at the

Two vertical poles $AB = 15 \ m$ and $CD = 10 \ m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $BC$ and $AD$,then the height of $P$ (in $m$) above the line $AC$ is:

The sum of the angles of elevation of the top of a tower from two points distant $a$ and $b$ from the base and in the same straight line with it is $90^{\circ}$. Then,the height of the tower is

$A$ person observes the top of a tower from a point $A$ on the ground. The angle of elevation of the tower from this point is $60^{\circ}$. He moves $60 \ m$ in the direction perpendicular to the line joining $A$ and the base of the tower to reach point $C$. The angle of elevation of the tower from point $C$ is $45^{\circ}$. Then, the height of the tower (in metres) is