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If \(\tan 2A = \cot(A - 18°)\), where 2A is an acute angle, find the value of A.

  • A.18°
  • B.24°
  • C.36°
  • D.48°

Answer: C

We use the identity \(\cot \theta = \tan(90° - \theta)\).

So, \(\cot(A - 18°) = \tan(90° - (A - 18°)) = \tan(108° - A)\).

The equation becomes \(\tan 2A = \tan(108° - A)\).

This implies \(2A = 108° - A\).

3A = 108°

A = 36°.

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