If the angles of a triangle are in the ratio $1 : 2 : 4$ , then prove that
$a^{2} b^{2} c^{2} = (b^{2} - a^{2}) (c^{2} - b^{2}) (a^{2} - c^{2})$ .

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Solution

$A + B + C = A + 2 A + 4 A = 7 A = π$
$∴ A = \frac{π}{7}$
Hence the angle are $A = \frac{π}{7}$ , $B = \frac{2 π}{7}$ , $C = \frac{4 π}{7}$
Now by sine Rule $\frac{a}{sin A} = λ$ (say)
$R . H . S = λ^{6} ({sin}^{2} B - {sin}^{2} A) ({sin}^{2} C - {sin}^{2} B) ({sin}^{2} A - {sin}^{2} C)$
$= λ^{6} [sin (B + A) . sin (B - A)] [sin (C + B) . sin (C - B)] [sin (A + C) . sin (A - C)]$
$= λ^{6} (sin \frac{3 π}{7} . sin \frac{π}{7}) (sin \frac{6 π}{7} . sin \frac{2 π}{7}) (sin \frac{5 π}{7} . sin \frac{3 π}{7})$
$sin \frac{6 π}{7} = sin (π - \frac{π}{7}) = sin \frac{π}{7}$
$sin \frac{5 π}{7} = sin (π - \frac{2 π}{7}) = sin \frac{2 π}{7}$
$∴ R . H . S = λ^{6} ({sin}^{2} \frac{π}{7} . {sin}^{2} \frac{2 π}{7} . {sin}^{2} \frac{4 π}{7})$
$= (λ^{2} {sin}^{2} A) (λ^{2} {sin}^{2} B) (λ^{2} {sin}^{2} C)$
$a^{2} b^{2} c^{2} = L . H . S .$

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