The upper part of a tree broken by the wind makes an angle of $30^\circ$ with the ground. The distance from the root to the point where the top of the tree touches the ground is $10 \, m$. What was the original height of the tree in meters?

From a point on the level ground, the angle of elevation of the top of a pole is $30^{\circ}$. On moving $20 \ m$ nearer to the pole, the angle of elevation becomes $45^{\circ}$. The height of the pole (in metres) 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

The angle of elevation of an object on a hill is observed from a certain point in the horizontal plane through its base to be $30^{\circ}$. After walking $120 \ m$ towards it on level ground, the angle of elevation is found to be $60^{\circ}$. Then the height of the object (in metres) is:

If the angles of elevation of the top of a tower from three collinear points $A, B$ and $C$ on a line leading to the foot of the tower are $30^o, 45^o$ and $60^o$ respectively,then the ratio $AB : BC$ 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