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

$A$ tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying on a horizontal line through the foot of the tower. Then $\frac{A B}{B C}$ is equal to:

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:

From the top of a hill $h$ metres high, the angles of depression of the top and the bottom of a pillar are $\alpha$ and $\beta$ respectively. The height (in metres) of the pillar is

The horizontal distance between a tower and a building is $10 \sqrt{3}$ units. If the angle of depression of the foot of the building from the top of the tower is $60^{\circ}$ and the angle of elevation of the top of the building from the foot of the tower is $30^{\circ}$,then the sum of the heights of the tower and the building is

Two poles standing on a horizontal ground are of heights $5 \, m$ and $10 \, m$ respectively. The line joining their tops makes an angle of $15^o$ with the ground. Then the distance (in $m$) between the poles is: