Question

For $A$ being acute, solve:

$\tan 2A=\cot (A-{18}^o)$.

Answer 1

Given,

$\tan 2A=\cot (A-{18}^o)$

or, $\cot ({90}^o-2A)=\cot (A-{18}^o)$

or, ${90}^o-2A=A-{18}^o$ [ Since $A$ being acute]

or, $3A={108}^o$

or, $A={36}^o$.