$A$ house subtends a right angle at the window of an opposite house. The angle of elevation of the window from the bottom of the first house is $30^{\circ}$. If the distance between the two houses is $6 \ m$,then the height of the first house is:

$A$ person observes the top of a tower from a point $A$ on the ground. The angle of elevation of the tower from this point is $60^{\circ}$. He moves $60 \ m$ in the direction perpendicular to the line joining $A$ and the base of the tower to reach point $C$. The angle of elevation of the tower from point $C$ is $45^{\circ}$. Then,the height of the tower (in metres) 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 = $

For a man,the angle of elevation of the highest point of the temple situated east of him is $60^\circ$. On walking $240 \ m$ to the north,the angle of elevation is reduced to $30^\circ$. The height of the temple is:

$A$ 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-

$A$ vertical pole subtends an angle $\tan ^{-1}\left(\frac{1}{2}\right)$ at a point $P$ on the ground. If the angles subtended by the upper half and the lower half of the pole at $P$ are respectively $\alpha$ and $\beta$,then $(\tan \alpha, \tan \beta)$ is equal to