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

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

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

Let a $$\Delta$$ABC having C = 90$$^\circ$$ and A = $$\theta$$

$${{\sin \theta } \over a} = {{\cos \theta } \over b} = {1 \over c}$$ ..... (i)

Also for triangle of reciprocals

$$\cos A = {{{{\left( {{1 \over c}} \right)}^2} + {{\left( {{1 \over b}} \right)}^2} - {{\left( {{1 \over a}} \right)}^2}} \over {2\left( {{1 \over c}} \right)\left( {{1 \over b}} \right)}}$$

$${1 \over {{c^2}}} + {1 \over {{{(c\cos \theta )}^2}}} = {1 \over {{{(c\sin \theta )}^2}}}$$

$$ \Rightarrow 1 + {\sec ^2}\theta = \cos e{c^2}\theta $$

$$ \Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {4{{\sin }^2}\theta {{\cos }^2}\theta }}$$

$$ \Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {{{\sin }^2}2\theta }}$$

$$ \Rightarrow 1 - {\cos ^2}2\theta = 4\cos 2\theta $$

$${\cos ^2}2\theta + 4\cos 2\theta - 1 = 0$$

$$\cos 2\theta = {{ - 4 \pm \sqrt {16 + 4} } \over 2}$$

$$\cos 2\theta = - 2 \pm \sqrt 5 $$

$$\cos 2\theta = \sqrt 5 - 2 = 1 - 2{\sin ^2}\theta $$

$$ \Rightarrow 2{\sin ^2}\theta = 3 - \sqrt 5 $$

$$ \Rightarrow {\sin ^2}\theta = {{3 - \sqrt 5 } \over 2}$$

$$ \Rightarrow \sin \theta = {{\sqrt 5 - 1} \over 2}$$