If $tan2A=cot (A-18^{o})$, where $2A$ is an angle, find the value of $A$.

Given: $tan2A=cot (A-18^{o})$, where $2A$ is an angle.

To do: To find the value of $A$.

Solution:

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

Now, As known that,

$tan\theta=cot( 90-\theta)$

$therefore cot( 90^{o}-2A)=cot( A-18^{o})$

$(90^{o}-2A)=(A-18^{o})$

$\Rightarrow 3A=90^{o}+18^{o}$

$\Rightarrow 3A=108^{o}$

$\Rightarrow A=\frac{108^{o}}{3}$

$\Rightarrow A=36^{o}$

Hence $A=36^{o}$
 

Updated on: 10-Oct-2022

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