The distance between the points $(a cos 48^{o}, 0)$ and $(0, a cos 12^{o})$ is d then $d^{2} - a^{2} = ?$

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Solution

given distance between the points $(a cos 48^{o}, 0)$ and $(0, cos 12^{o})$ is $d$
$\sqrt{{(a cos 48^{o})}^{2} + {(a cos 12^{o})}^{2}} = d$
$\Rightarrow a^{2} ({cos}^{2} 48^{o} + {cos}^{2} 12^{o}) = d^{2} \Rightarrow a^{2} ({(cos (60 - 12))}^{2} + {cos}^{2} 12^{o}) = d^{2}$
$\Rightarrow a^{2} ⎛ ⎝ {(\frac{1}{2} cos 12^{o} + \frac{\sqrt{3}}{2} sin 12^{o})}^{2} + {cos}^{2} 12^{o} ⎞ ⎠ = d^{2} \Rightarrow a^{2} (\frac{5}{4} {cos}^{2} 12^{o} + \frac{3}{4} {sin}^{2} 12^{o} + \frac{\sqrt{3}}{2} sin 12^{o} cos 12^{o}) = d^{2}$
$\Rightarrow a^{2} (\frac{5}{4} - \frac{1}{2} {sin}^{2} 12^{o} + \frac{\sqrt{3}}{4} sin 24^{o}) = d^{2} \Rightarrow a^{2} (\frac{5}{4} - \frac{1}{4} (1 - cos 24^{o}) + \frac{\sqrt{3}}{4} sin 24^{o}) = d^{2}$
$\Rightarrow a^{2} (1 + \frac{1}{2} (\frac{1}{2} cos 24^{o} + \frac{\sqrt{3}}{2} sin 24^{o})) = d^{2} \Rightarrow a^{2} (1 + \frac{1}{2} cos (60^{o} - 24^{o})) = d^{2}$
$\Rightarrow d^{2} - a^{2} = \frac{a^{2}}{2} cos 48^{o}$

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