The sum of the 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

$A$ tower stands at the centre of a circular park. $A$ and $B$ are two points on the boundary of the park such that $AB = a$ subtends an angle of $60^{\circ}$ at the foot of the tower,and the angle of elevation of the top of the tower from $A$ or $B$ is $30^{\circ}$. The height of the tower is

From the top of a lighthouse $30 \ m$ high with its base at sea level,the angle of depression of a boat is $15^\circ$. The distance of the boat from the foot of the lighthouse 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 = $

The angle of elevation of the top of a tower from a point $A$ due north of it is $\alpha$ and from a point $B$ at a distance of $9$ units due west of $A$ is $\cos^{-1}\left(\frac{3}{\sqrt{13}}\right)$. If the distance of the point $B$ from the tower is $15$ units,then $\cot \alpha$ is equal to.

The horizontal distance between two towers is $60 \ m$. The angle of depression of the top of the first tower as seen from the top of the second tower is $30^\circ$. If the height of the second tower is $150 \ m$,then the height of the first tower is: