$A$ 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:

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

Two vertical poles of equal heights are $120 \, m$ apart. On the line joining their bottoms,$A$ and $B$ are two points. The angle of elevation of the top of one pole from $A$ is $45^\circ$ and that of the other pole from $B$ is also $45^\circ$. If $AB = 30 \, m$,then the height of each pole is.....$m$.

From a $60 \ m$ high tower,the angles of depression of the top and bottom of a house are $\alpha$ and $\beta$ respectively. If the height of the house is $\frac{60 \sin(\beta - \alpha)}{x}$,then $x =$

The sum of angles of elevation of the top of a tower from two points distant $a$ and $b$ from the base and in the same straight line with it is $90^{\circ}$. Then, the height of the tower is

$A$ tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying on a horizontal line through the foot of the tower. Then $\frac{A B}{B C}$ is equal to: