$P$ is a point on the segment joining the feet of two vertical poles of heights $a$ and $b$. The angles of elevation of the tops of the poles from $P$ are $45^{\circ}$ each. Then, the square of the distance between the tops of the poles is

The angle of elevation of the top $P$ of a vertical tower $PQ$ of height $10$ from a point $A$ on the horizontal ground is $45^{\circ}$. Let $R$ be a point on $AQ$ and from a point $B$,vertically above $R$,the angle of elevation of $P$ is $60^{\circ}$. If $\angle BAQ = 30^{\circ}$,$AB = d$ and the area of the trapezium $PQRB$ is $\alpha$,then the ordered pair $(d, \alpha)$ is.

The angle of elevation of the top of a pillar at a point $A$ on the ground is $15^\circ$. On walking $40 \ m$ towards the pillar,the angle becomes $30^\circ$. The height of the pillar is $... \ m$.

The shadow of a tower standing on a level ground is found to be $60 \ m$ longer when the Sun's altitude is $30^{\circ}$ than when it is $45^{\circ}$. The height of the tower is

$ABC$ is a triangular park with $AB = AC = 100 \text{ metres}$. $A$ vertical tower of height $h$ is situated at the mid-point $P$ of $BC$. If the angles of elevation of the top of the tower $Q$ at $A$ and $B$ are $\cot^{-1}(3\sqrt{2})$ and $\csc^{-1}(2\sqrt{2})$ respectively,then the height of the tower (in metres) is

$A$ tower $50 \ m$ high stands on the top of a mount. From a point on the ground,the angles of elevation of the top and bottom of the tower are found to be $75^o$ and $60^o$ respectively. The height of the mount is: