The horizontal distance between two towers is $60 \ m$. The angle of depression of the top of the first tower as seen from the top of the second tower is $30^\circ$. If the height of the second tower is $150 \ m$,then the height of the first tower is:

The length of the shadow of a pole inclined at $10^\circ$ to the vertical towards the sun is $2.05 \text{ m}$,when the elevation of the sun is $38^\circ$. The length of the pole 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

$A$ man is walking towards a vertical pillar in a straight path,at a uniform speed. At a certain point $A$ on the path,he observes that the angle of elevation of the top of the pillar is $30^o$. After walking for $10 \text{ minutes}$ from $A$ in the same direction,at a point $B$,he observes that the angle of elevation of the top of the pillar is $60^o$. Then the time taken (in minutes) by him,from $B$ to reach the pillar,is:

$A$ pole stands vertically inside a triangular park $\Delta ABC$. Let the angle of elevation of the top of the pole from each corner of the park be $\frac{\pi}{3}$. If the radius of the circumcircle of $\Delta ABC$ is $2$,then the height of the pole is equal to:

From the bottom of a pole of height $h,$ the angle of elevation of the top of a tower is $\alpha$ and the pole subtends an angle $\beta$ at the top of the tower. The height of the tower is