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

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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 $36^{\circ}$.

Updated on 10-Oct-2022 13:22:36