The sum of angles of elevation of the top of a tower from two points distant $a$ and $b$ from the base and in the same straight line with it is $90^{\circ}$. Then,the height of the tower is

From the base of a pole of height $20 \text{ m}$,the angle of elevation of the top of a tower is $60^{\circ}$. The pole subtends an angle of $30^{\circ}$ at the top of the tower. Then the height of the tower is

The angle of elevation of an object on a hill is observed from a certain point in the horizontal plane through its base to be $30^{\circ}$. After walking $120 \ m$ towards it on level ground,the angle of elevation is found to be $60^{\circ}$. Then the height of the object (in metres) is:

An aeroplane flying with uniform speed horizontally $1 \ km$ above the ground is observed at an elevation of $60^{\circ}$. After $10 \ s$,if the elevation is observed to be $30^{\circ}$,then the speed of the plane (in $km/h$) is

$A$ spherical gas balloon of radius $16 \ m$ subtends an angle $60^{\circ}$ at the eye of the observer $A$,while the angle of elevation of its center from the eye of $A$ is $75^{\circ}$. Then the height (in $m$) of the top most point of the balloon from the level of the observer's eye 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 = $