$A$ pole stands vertically inside a triangular park $ABC$. If the angle of elevation of the top of the pole from each corner of the park is the same,then in the $\Delta ABC$,the foot of the pole is at the

$A$ tower $T_1$ of height $60 \, m$ is located exactly opposite to a tower $T_2$ of height $80 \, m$ on a straight road. From the top of $T_1$,if the angle of depression of the foot of $T_2$ is twice the angle of elevation of the top of $T_2$,then the width (in $m$) of the road between the feet of the towers $T_1$ and $T_2$ 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

$A$ vertical pole fixed to the horizontal ground is divided in the ratio $3:7$ by a mark on it,with the lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground $18 \ m$ away from the base of the pole,then the height of the pole (in $meters$) is:

From the base of a pole of height $20 \text{ m}$,the angle of elevation of the top of a tower is $60^{\circ}$. The pole subtends an angle of $30^{\circ}$ at the top of the tower. Then the height of the tower is

If a flagstaff of $6 \ m$ high placed on the top of a tower throws a shadow of $2\sqrt{3} \ m$ along the ground,then the angle (in degrees) that the sun makes with the ground is.......$^o$