In $\Delta ABC$,$m\angle B = 90^\circ$,$m\angle C = 30^\circ$ and $AC = 30 \text{ cm}$,then $AB = \ldots \text{ cm}$.

The angle of elevation of the top of a tower $30 \, m$ high from the foot of another tower in the same plane is $60^{\circ}$ and the angle of elevation of the top of the second tower from the foot of the first tower is $30^{\circ}$. Find the distance between the two towers and also the height of the other tower.

$A$ pole $6\, m$ high casts a shadow $2 \sqrt{3}\, m$ long on the ground. Then,the Sun's elevation is: (in $^{\circ}$)

The angle of elevation of the top of a vertical tower from a point on the ground is $60^{\circ}$. From another point $10 \, m$ vertically above the first,its angle of elevation is $45^{\circ}$. Find the height of the tower.

The angle of elevation of a cloud from a point $h$ metres above the surface of a lake is $\theta$ and the angle of depression of its reflection in the lake is $\phi$. Prove that the height of the cloud above the lake is $h\left(\frac{\tan \phi+\tan \theta}{\tan \phi-\tan \theta}\right).$

Write 'True' or 'False' and justify your answer.
If the height of a tower and the distance of the point of observation from its foot,both,are increased by $10 \%$,then the angle of elevation of its top remains unchanged.