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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 $A=36^{o}$
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