$A$ man from the top of a $100 \ m$ high tower sees a car moving towards the tower at an angle of depression of $30^\circ$. After some time,the angle of depression becomes $60^\circ$. The distance (in meters) travelled by the car during the time 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:

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:

$A$ 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

The angle of elevation of a tower at a point distant $d$ meters from its base is $30^\circ$. If the tower is $20$ meters high,then the value of $d$ is

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object, the angle of elevation is found to be $\beta$. Then the height (in metres) of the object is