The angle of elevation of a stationary cloud from a point $2500 \, m$ above a lake is $15^\circ$ and the angle of depression of its reflection in the lake is $45^\circ$. The height of the cloud above the lake level is

The angle of elevation of a stationary cloud from a point $2500 \ m$ above a lake is $15^{\circ}$ and from the same point the angle of depression of its reflection in the lake is $45^{\circ}$. The height (in metres) of the cloud above the lake,given that $\cot 15^{\circ}=2+\sqrt{3}$,is

$AB$ is a vertical pole resting at the end $A$ on the level ground. $P$ is a point on the level ground such that $AP = 3 \, AB$. If $C$ is the mid-point of $AB$ and $CB$ subtends an angle $\beta$ at $P$,the value of $\tan \beta$ is

$A$ tower subtends angles $\alpha, 2\alpha, 3\alpha$ respectively at points $A, B$ and $C$,all lying on a horizontal line through the foot of the tower. Then $AB/BC = $

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:

$A$ person is standing on a tower of height $15(\sqrt{3} + 1) \ m$ and observing a car coming towards the tower. He observes that the angle of depression changes from $30^\circ$ to $45^\circ$ in $3 \ s$. What is the speed of the car in $km/hr$?