If $ \tan 2 A=\cot \left(A-18^{\circ}\right) $, where $ 2 A $ is an acute angle, find the value of $ A $.
Given:
\( \tan 2 A=\cot \left(A-18^{\circ}\right) \), where \( 2 A \) is an acute angle.
To do:
We have to find the value of \( A \).
Solution:
We know that,
$\cot (90^{\circ}- \theta) = tan\ \theta$
Therefore,
$\tan 2 A=\cot\left(A-18^{\circ}\right)$
$\cot (90^{\circ}- 2A)=\cot\left(A-18^{\circ}\right)$
Comparing on both sides, we get,
$90^{\circ}-2A =A-18^{\circ}$
$2A+A=90^{\circ}+18^{\circ}$
$3A=108^{\circ}$
$A=\frac{108^{\circ}}{3}$
$A=36^{\circ}$
The value of \( A \) is $36^{\circ}$.
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