The angle of elevation of the top of a vertical tower from a point $A$,due east of it,is $45^\circ$. The angle of elevation of the top of the same tower from a point $B$,due south of $A$,is $30^\circ$. If the distance between $A$ and $B$ is $54\sqrt{2} \, \text{m}$,then the height of the tower (in metres) is:

The angle of elevation of the top of a $TV$ tower from three points $A$,$B$ and $C$ in a straight line through the foot of the tower are $\alpha, 2 \alpha$ and $3 \alpha$ respectively. If $AB = a$,then the height of the tower is

$A$ 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:

$P$ 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

$A$ vertical pole fixed to the horizontal ground is divided in the ratio $3:7$ by a mark on it,with the lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 \ m$ away from the base of the pole,then the height of the pole (in $meters$) is:

$A$ man is walking towards a vertical pillar in a straight path,at a uniform speed. At a certain point $A$ on the path,he observes that the angle of elevation of the top of the pillar is $30^o$. After walking for $10 \text{ minutes}$ from $A$ in the same direction,at a point $B$,he observes that the angle of elevation of the top of the pillar is $60^o$. Then the time taken (in minutes) by him,from $B$ to reach the pillar,is: