Two men are on the opposite sides of a tower. They measure the angles of elevation of the top of the tower as $45^{\circ}$ and $30^{\circ}$ respectively. If the height of the tower is $40 \, m,$ find the distance between the men.

$A$ 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:

An observer on the top of a tree finds the angle of depression of a car moving towards the tree to be $30^\circ$. After $3 \text{ minutes}$,this angle becomes $60^\circ$. After how much more time will the car reach the tree? (in minutes)

The shadow of a tower standing on a level plane is found to be $100 \ m$ longer when the sun's angle of elevation is $30^{\circ}$ than when it is $45^{\circ}$. The height of the tower is:

From a point on the level ground, the angle of elevation of the top of a pole is $30^{\circ}$. On moving $20 \ m$ nearer to the pole, the angle of elevation becomes $45^{\circ}$. The height of the pole (in metres) 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.