$$\tan \theta = {1 \over 2}$$

$$\tan (\theta + \alpha ) = {x \over b},\tan \alpha = {x \over {a + b}}$$

$$ \Rightarrow {{{1 \over 2} + {x \over {a + b}}} \over {1 - {1 \over 2} \times {x \over {a + b}}}} = {x \over b}$$

$$ \Rightarrow {x^2} - 2ax + ab + {b^2} = 0$$

$${{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}$$

As A, B, C are angles of triangle.

A + B + C = $$\pi$$

A = $$\pi$$ $$-$$ (B + C) ...... (1)

Similarly sinB = sin(A + C) ..... (2)

From (1) and (2)

$${{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \over {\sin (C - B)}}$$

$$\sin (C + B).\sin (C - B) = \sin (A - C)\sin (A + C)$$

$${\sin ^2}C - {\sin ^2}B = {\sin ^2}A - {\sin ^2}C$$

$$\because$$ $$\{ \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\} $$

$$2{\sin ^2}C = {\sin ^2}A + {\sin ^2}B$$

By sine rule

$$2{c^2} = {a^2} + {b^2}$$

$$\Rightarrow$$ b², c² and a² are in A.P.

From figure,

$$\sin \beta = {1 \over {\sqrt 5 }}$$

$$\therefore$$ $$\tan \beta = {1 \over 2}$$

$$\tan (\alpha + \beta ) = 2$$

$${{\tan \alpha + \tan \beta )} \over {1 - \tan \alpha .\tan \beta }} = 2$$

$${{\tan \alpha + {1 \over 2}} \over {1 - \tan \alpha \left( {{1 \over 2}} \right)}} = 2$$

$$\tan \alpha = {3 \over 4}$$

$$\alpha = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$

O $$\to$$ centre of sphere

P, Q $$\to$$ point of contact of tangents from A

Let T be top most point of balloon & R be foot of perpendicular from O to ground.

From triangle OAP, OA = 16cosec30$$^\circ$$ = 32

From triangle ABO, OR = OA sin75$$^\circ$$ = $$32{{\left( {\sqrt 3 + 1} \right)} \over {2\sqrt 2 }}$$

So level of top most point = OR + OT

$$ = 8\left( {\sqrt 6 + \sqrt 2 + 2} \right)$$