$$\eqalign{ & \tan {15^ \circ } = \frac{{60}}{x} \cr & \Rightarrow \,\,x = 60\cot {15^ \circ } \cr & = 60\left[ {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right] \cr} $$

Here, $$\frac{{BD}}{{AB}} = \frac{{DI}}{{AI}}$$
$$\eqalign{ & \Rightarrow \,\,\frac{{\frac{a}{2}}}{c} = \frac{1}{1} \cr & \therefore \,\,a = 2c. \cr} $$
Similarly, $$b = 2c.$$
$$\eqalign{ & \therefore \,\,\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{4{c^2} + {c^2} - 4{c^2}}}{{2 \cdot 2c \cdot c}} \cr & \cos A = \frac{1}{4}. \cr} $$

$$\eqalign{ & HD = BD \cdot \tan \angle EBC = c\cos B \cdot \tan \left( {{{90}^ \circ } - C} \right) \cr & HD = \frac{{c\cos B \cdot \cos C}}{{\sin C}} = 2R\cos B\cos C \cr & HD = \frac{{2R\cos A\cos B\cos C}}{{\cos A}}. \cr} $$
Similarly for others. So, the ratio of the distances of the orthocentre from the sides $$ = \frac{1}{{\cos A}}:\frac{1}{{\cos B}}:\frac{1}{{\cos C}}.$$

$$\eqalign{ & c\left( {1 + \cos A} \right) + a\left( {1 + \cos C} \right) = 3b \cr & \Rightarrow \,\,a + c + \left( {c\cos A + a\cos C} \right) = 3b \cr & \Rightarrow \,\,a + c + b = 3b. \cr} $$

$$\eqalign{ & {\text{In }}\Delta \,ADB,\tan {45^ \circ } = \frac{{AD}}{{BD}} = \frac{{AD}}{{21}} \cr & AD = 21\,m \cr & {\text{In }}\Delta \,ADC,\tan {60^ \circ } = \frac{{AD}}{x} \cr & x = \frac{{21}}{{\sqrt 3 }} = 7\sqrt 3 \,m \cr} $$

If $$a\,{\cos ^2}\left( {\frac{C}{2}} \right) + c\,{\cos ^2}\left( {\frac{A}{2}} \right) = \frac{{3b}}{2}$$
$$\eqalign{ & a\left[ {\cos C + 1} \right] + c\left[ {\cos A + 1} \right] = 3b \cr & \left( {a + c} \right) + \left( {a\cos C + c\cos B} \right) = 3b \cr} $$
$$a + c + b = 3b\,\,{\text{or }}\,a + c = 2b\,\,{\text{or}}$$       $$a, b, c$$   are in A.P.

$$QS : SR = PQ : PR$$     (as bisector of an angle, in a triangle, divides the opposite side in the same ratio as the sides containing the angle.)

$$\eqalign{ & a = c\sin \theta ,b = c\cos \theta \cr & \Rightarrow {\left( {\frac{c}{a} + \frac{c}{b}} \right)^2} \cr & = {\left( {\frac{1}{{\sin \theta }} + \frac{1}{{\cos \theta }}} \right)^2} = \frac{{4\left( {1 + \sin 2\theta } \right)}}{{{{\sin }^2}2\theta }} \cr & = 4\left( {\frac{1}{{{{\sin }^2} 2\theta }} + \frac{1}{{\sin 2\theta }}} \right),{\text{where }}\,0 < \theta < \frac{\pi }{2} \cr & \Rightarrow {\left( {\frac{c}{a} + \frac{c}{b}} \right)^2}_{\min } = 8,{\text{when }}\,2\theta = {90^ \circ }. \cr} $$

Here, $$\vartriangle = {a^2} - {\left( {b - c} \right)^2}$$
$$\eqalign{ & \Rightarrow \,\,\vartriangle = \left( {2s - 2c} \right)\left( {2s - 2b} \right) = 4\left( {s - b} \right)\left( {s - c} \right) \cr & \therefore \,\,\frac{1}{4} = \frac{{\left( {s - b} \right)\left( {s - c} \right)}}{\vartriangle } = \tan \frac{A}{2} \cr & \Rightarrow \,\,\tan A = \frac{{2\tan \frac{A}{2}}}{{1 - {{\tan }^2}\frac{A}{2}}} = \frac{{2 \times \frac{1}{4}}}{{1 - {{\left( {\frac{1}{4}} \right)}^2}}}. \cr} $$

Let $$QT$$  be the tower of height $$\left( h \right)$$  in $$\Delta \,PRS.$$
Now, each triangle $$QPR, QRS, QSP$$    are equilateral.
Thus, $$QP = QS = QR = a.$$
In $$\Delta \,QTP,$$
$$\eqalign{ & Q{P^2} = Q{T^2} + P{T^2} \cr & \Rightarrow {a^2} = {h^2} + {\left( {\frac{a}{2}\sec {{30}^ \circ }} \right)^2} \cr} $$

$$\eqalign{ & \Rightarrow {a^2} = {h^2} + \frac{{{a^2}}}{4} \cdot \frac{4}{3} \cr & \Rightarrow {a^2} = {h^2} + \frac{{{a^2}}}{3} \cr & \Rightarrow {a^2} - \frac{{{a^2}}}{3} = {h^2} \cr & \Rightarrow \frac{{3{a^2} - {a^2}}}{3} = {h^2} \cr & \therefore 2{a^2} = 3{h^2} \cr} $$