An observer $1.5 \ m$ tall is $20.5 \ m$ away from a tower $22 \ m$ high. Determine the angle of elevation of the top of the tower from the eye of the observer. (in $^{\circ}$)

$A$ straight highway leads to the foot of a tower. $A$ man standing at the top of the tower observes a car at an angle of depression of $30^{\circ}$,which is approaching the foot of the tower with a uniform speed. Six seconds later,the angle of depression of the car is found to be $60^{\circ}$. Find the time taken by the car to reach the foot of the tower from this point (in $seconds$).

From the top of a tower,the angles of depression of points $A$ and $B$ in the same direction of the tower are found to be $\theta$ and $(90^\circ - \theta)$ respectively. If $B$ is nearer the tower than $A$ and $AB = a$,then prove that the height of the tower is $\frac{a \tan \theta}{1 - \tan^2 \theta}$.

From a point $P$ on the ground,the angle of elevation of the top of a $10 \, m$ tall building is $30^{\circ}$. $A$ flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from $P$ is $45^{\circ}$. Find the length of the flagstaff and the distance of the building from the point $P$. (You may take $\sqrt{3} = 1.732$)

As observed from the top of a lighthouse,the angles of depression of two ships $P$ and $Q$ anchored in the sea on the same side are found to be $35^{\circ}$ and $50^{\circ}$ respectively. Then,from the lighthouse:

The angle of elevation of the top of the temple on the bank of the river from a point exactly opposite to the temple on the other bank of the river is $30^{\circ}$. If the height of the temple is $20 \ m$,then the breadth of the river is $\ldots \ldots \ldots \ldots \ m$.