Let the angles $$A, B, C$$ of a triangle $$ABC$$ be in A.P. and let $$b:c = \sqrt 3 :\sqrt 2 $$. Find the angle $$A$$.

(i) $$PQ$$ is a vertical tower. $$P$$ is the foot and $$Q$$ is the top of the tower. $$A, B, C$$ are three points in the horizontal plane through $$P$$. The angles of elevation of $$Q$$ from $$A$$, $$B$$, $$C$$ are equal, and each is equal to $$\theta $$. The sides of the triangle $$ABC$$ are $$a, b, c$$; and the area of the triangle $$ABC$$ is $$\Delta $$. Show that the height of the tower is $${{abc\tan \theta } \over {4\Delta }}$$.

(ii) $$AB$$ is vertical pole. The end $$A$$ is on the level ground. $$C$$ is the middle point of $$AB$$. $$P$$ is a point on the level ground. The portion $$CB$$ subtends an angle $$\beta $$ at $$P$$. If $$AP = n\,AB,$$ then show that tan$$\beta $$ $$ = {n \over {2{n^2} + 1}}$$

$$ABC$$ is a triangle with $$AB=AC$$. $$D$$ is any point on the side $$BC$$. $$E$$ and $$F$$ are points on the side $$AB$$ and $$AC$$, respectively, such that $$DE$$ is parallel to $$AC$$, and $$DF$$ is parallel to $$AB$$. Prove that $$$DF + FA + AE + ED = AB + AC$$$

$$ABC$$ is a triangle. $$D$$ is the middle point of $$BC$$. If $$AD$$ is perpendicular to $$AC$$, then prove that $$$\cos A\,\cos C = {{2\left( {{c^2} - {a^2}} \right)} \over {3ac}}$$$