The angle of elevation of a tower from a point $A$ due south of it is $30^\circ$ and from a point $B$ due west of it is $45^\circ$. If the height of the tower is $100 \ m$,then $AB = \dots \ m$.

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object, the angle of elevation is found to be $\beta$. Then the height (in metres) of the object 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

$A$ horizontal park is in the shape of a triangle $OAB$ with $AB = 16$. $A$ vertical lamp post $OP$ is erected at the point $O$ such that $\angle PAO = \angle PBO = 15^{\circ}$ and $\angle PCO = 45^{\circ}$,where $C$ is the midpoint of $AB$. Then $(OP)^{2}$ is equal to.

From the top of a hill $h$ metres high, the angles of depression of the top and the bottom of a pillar are $\alpha$ and $\beta$ respectively. The height (in metres) of the pillar is

$A$ tower subtends an angle of $30^\circ$ at a point distant $d$ from the foot of the tower and on the same level as the foot of the tower. At a second point $h$ vertically above the first,the angle of depression of the foot of the tower is $60^\circ$. The height of the tower is: