In the figure,$\theta_1+\theta_2=\frac{\pi}{2}$ and $\sqrt{3}(BE)=4(AB)$. If the area of $\triangle CAB$ is $2\sqrt{3}-3 \text{ unit}^2$,when $\frac{\theta_2}{\theta_1}$ is the largest,then the perimeter (in unit) of $\triangle CED$ is equal to $...........$.

The angular elevation of a tower $CD$ at a point $A$ due south of it is $60^\circ$ and at a point $B$ due west of $A$,the elevation is $30^\circ$. If $AB = 3 \, km$,the height of the tower is

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:

$A$ vertical pole (more than $100 \, ft$ high) consists of two portions,the lower being one-third of the whole. If the upper portion subtends an angle $\tan^{-1}(\frac{1}{2})$ at a point in a horizontal plane through the foot of the pole and at a distance of $40 \, ft$ from it,then the height of the pole is......$ft$.

$A$ 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

$A$ tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying on a horizontal line through the foot of the tower. Then $\frac{A B}{B C}$ is equal to: