Solution: Relation between polar Co-ordinate system and Cartesian Co-ordinate system \(\displaystyle{x}={r}{\cos{{0}}}{y}={r}{\sin{{0}}}\)

\({r}=\sqrt{{{x}^{{2}}+{y}^{{2}}}}\)

given that - \(\displaystyle{\frac{{{x}^{{2}}}}{{{9}}}}-{\frac{{{y}^{{2}}}}{{{16}}}}={25}\) put \(x = r \cos 0\), \(y = e \sin 0\)

\(\Rightarrow \frac{r^2 cos^2 0}{0} - \frac{r^2 sin^2 0}{16} = 25\)

\(\Rightarrow r^2 [\frac{16 cos^2 0 - 9 sin^2 0}{177}] = 25\)

\(\Rightarrow r^2 = \frac{25 \times 177 }{[(7 cos)^2 - (3 sin)^2]} [a^2 - h^2 = (a - b) (a + b)]\)

\(\Rightarrow r^2 = \frac{25 \times 177}{(7 cos 0 - 3 sin 0)(4 cos 0 + 3 sin 0)}\)

\(\Rightarrow r^2 = \frac{25 \times 177}{5[(\frac{7}{5} cos 0 - \frac{3}{5} sin 0)] 5 [(\frac{7}{5} cos 0 + \frac{3}{5} sin 0)]}\)

\(\Rightarrow r^2 \frac{177}{(\frac{7}{5} cos 0 - \frac{3}{5} sin 0)(\frac{7}{5} cos 0 + \frac{3}{5} sin 0)} [sin(53^{\circ}) = \frac{3}{5}]\)

\(\Rightarrow r^2 = \frac{177}{(sin 53^{\circ} cos 0 - cos 53^{\circ} sin 0)(sin 53^{\circ} cos 0 + cos 53^{\circ} sin 0)}\) formula

\(\displaystyle{\sin{{\left({A}+{B}\right)}}}={\sin{{A}}}{\cos{{B}}}+{\cos{{A}}}{\sin{{B}}}\)

\({\sin{{\left({A}-{B}\right)}}}={\sin{{A}}}{\cos{{B}}}-{\cos{{A}}}{\sin{{B}}}\)

\(\Rightarrow{r}^{{2}}{\frac{{{177}}}{{{\left[{\sin{{\left({53}-{0}\right)}}}\right]}{\left[{\sin{{\left({53}+{0}\right)}}}\right]}}}}\)

\(\Rightarrow{r}^{{2}}={\frac{{{177}}}{{{\left[{\sin{{\left({53}-{0}\right)}}}\right]}{\left[{\sin{{\left({0}+{53}\right)}}}\right]}}}}\)

Finally answer is \(\displaystyle\Rightarrow{r}^{{2}}={\frac{{{177}}}{{{\left[{\sin{{\left({53}-{0}\right)}}}\right]}{\left[{\sin{{\left({0}+{53}\right)}}}\right]}}}}\)