$A$ 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 = $

$A$ 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-

$A$ tower $50 \ m$ high stands on the top of a mount. From a point on the ground,the angles of elevation of the top and bottom of the tower are found to be $75^o$ and $60^o$ respectively. The height of the mount is:

The angle of elevation of a cloud $C$ from a point $P$,$200 \ m$ above a still lake is $30^{\circ}$. If the angle of depression of the image of $C$ in the lake from the point $P$ is $60^{\circ}$,then $PC$ (in $m$) is equal to

Let $10$ vertical poles standing at equal distances on a straight line,subtend the same angle of elevation $\alpha$ at a point $O$ on this line and all the poles are on the same side of $O$. If the height of the longest pole is $h$ and the distance of the foot of the smallest pole from $O$ is $a$,then the distance between two consecutive poles is:

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