The angle of elevation of a point situated at a distance of $70 \ m$ from the base of a tower is $45^\circ$. The height of the tower is ..... $m$.

$A$ man is observing,from the top of a tower,a boat speeding towards the tower from a certain point $A$,with uniform speed. At that point,the angle of depression of the boat from the man's eye is $30^{\circ}$ (ignore the man's height). After sailing for $20 \text{ seconds}$ towards the base of the tower (which is at the level of water),the boat reaches a point $B$,where the angle of depression is $45^{\circ}$. Then,the time taken (in seconds) by the boat from $B$ to reach the base of the tower is:

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

Consider a triangular plot $ABC$ with sides $AB = 7 \ m$,$BC = 5 \ m$,and $CA = 6 \ m$. $A$ vertical lamp-post at the midpoint $D$ of $AC$ subtends an angle $30^o$ at $B$. The height (in $m$) of the lamp-post 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.