Question

If tan2A = cot (A – 18°), where 2A is an acute angle, then A = 

A) 6° 

B) 18° 

C) 36° 

D) 54°

Answer 1

Correct option is: C) 36°

We have tan 2 A = cot (A- \(18^\circ\)),

where 2A is an acute angle.

= Cot (\(90^\circ\) - 2A) = cot (A - \(18^\circ\)) (\(\because\) 0 < 2A < \(90^\circ\) = \(90^\circ\) - 2 A > 0 & cot ( \(90^\circ\) - \(\theta\)) = tan \(\theta\))

= \(90^\circ\) - 2A = A - \(18^\circ\)

= 3A = \(90^\circ\) + \(18^\circ\) = \(108^\circ\)

= A = \(\frac {108^\circ}3\) = \(36^\circ\)

Answer 2

Correct option is: C) 36°