$P$ 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

$A$ vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles subtended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$, then $(\tan \alpha, \tan \beta)$ is equal to

$A$ tree is broken by wind,its upper part touches the ground at a point $10 \ m$ from the foot of the tree and makes an angle of $45^\circ$ with the ground. The total length of the tree is:

$A$ tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying on a horizontal line through the foot of the tower. Then $\frac{A B}{B C}$ is equal to:

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.

$A$ house subtends a right angle at the window of an opposite house and the angle of elevation of the window from the bottom of the first house is $60^\circ$. If the distance between the two houses is $6 \ m$,then the height of the first house is: