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

$A$ ladder $5 \ m$ long leans against a vertical wall. The bottom of the ladder is $3 \ m$ from the wall. If the bottom of the ladder is pulled $1 \ m$ farther from the wall,how much does the top of the ladder slide down the wall?

The shadow of a tower of height $(1 + \sqrt{3}) \text{ m}$ standing on the ground is found to be $2 \text{ m}$ longer when the sun's elevation is $30^{\circ}$ than when the sun's elevation was $...^{\circ}$.

$AB$ is a vertical pole with $B$ at the ground level and $A$ at the top. $A$ man finds that the angle of elevation of the point $A$ from a certain point $C$ on the ground is $60^{\circ}$. He moves away from the pole along the line $BC$ to a point $D$ such that $CD = 7 \ m$. From $D$,the angle of elevation of the point $A$ is $45^{\circ}$. Then the height of the pole 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

$A$ sphere with centre $O$ sits on the top of a pole as shown in the figure. An observer on the ground is at a distance $50 \ m$ from the foot of the pole. She notes that the angles of elevation from the observer to points $P$ and $Q$ on the sphere are $30^{\circ}$ and $60^{\circ}$,respectively. Then,the radius of the sphere in metres is