An aeroplane flying horizontally $1 \ km$ above the ground is observed at an elevation of $60^\circ$ and after $10$ seconds the elevation is observed to be $30^\circ$. The uniform speed of the aeroplane in $km/h$ is

$A$ vertical pole of height $h$ stands on a building of height $H$. The angle of elevation of the top of the building from a point on the ground,which is $5$ units away from the foot of the building,is $\alpha$. The pole subtends an angle $\beta$ at the same point on the ground. If $2 \tan \beta \cot \alpha = 1$,then-

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:

$A$ 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 a point $a$ metre above a lake,the angle of elevation of a cloud is $\alpha$ and the angle of depression of its reflection is $\beta$. The height of the cloud is

$A$ person standing on the top of a building of height $60 \sqrt{3}$ feet observed the top of a tower to lie at an elevation of $45^{\circ}$. That person descended to the bottom of the building and found that the top of the same tower is now at an angle of elevation of $60^{\circ}$. The height of the tower (in feet) is