At a point on the ground,the angle of elevation of a tower is such that its cotangent is $3/5$. On walking $32 \, m$ towards the tower,the cotangent of the angle of elevation becomes $2/5$. The height of the tower is .... $m$

$P$ 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

Two parallel towers $A$ and $B$ of different heights are at some distance $d$ on the same level ground. If the angle of elevation of a point $P$ at $20\,m$ height on tower $B$ from a point $Q$ at $10\,m$ height on tower $A$ is $\theta$,and this angle is equal to half the angle of elevation of point $R$ at $50\,m$ height on tower $A$ from point $P$ on tower $B$,then $\theta$ is equal to....$^o$

Two poles,$AB$ of length $a$ metres and $CD$ of length $a+b$ $(b \neq a)$ metres are erected at the same horizontal level with bases at $B$ and $D$. If $BD=x$ and $\tan \angle ACB = \frac{1}{2}$,then:

The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be $45^{\circ}$. After walking a distance of $80 \ m$ towards the top,up a slope inclined at an angle of $30^{\circ}$ to the horizontal plane,the angle of elevation of the top of the hill becomes $75^{\circ}$. Then the height of the hill (in meters) is

The angle of depression of a ship from the top of a tower $30 \, m$ high is $60^\circ$. Find the distance of the ship from the base of the tower.