$A$ 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 a cliff at a point $A$ on the ground and a point $B$,$100 \ m$ vertically above $A$,are $\alpha$ and $\beta$ respectively. The height of the cliff is

The angle of elevation of the top of the tower observed from each of the three points $A, B, C$ on the ground,forming a triangle,is the same angle $\alpha$. If $R$ is the circum-radius of the $\Delta ABC$,then the height of the tower is

$P$ is a point on the segment joining the feet of two vertical poles of heights $a$ and $b$. The angles of elevation of the tops of the poles from $P$ are $45^{\circ}$ each. Then, the square of the distance between the tops of the poles is

The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be $45^o$ from a point $A$ on the plane. Let $B$ be a point $30 \, m$ vertically above the point $A$. If the angle of elevation of the top of the tower from $B$ is $30^o$,then the distance (in $m$) of the foot of the tower from the point $A$ is:

$A$ person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is $60^\circ$. When he moves $40 \ m$ away from the bank,he finds the angle to be $30^\circ$. The breadth of the river is.....$m$