If tan 2A = cot(A - 18^{circ}),where 2A is an | vedclass

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(D) Given the equation: $	an 2A = \cot(A - 18^{\circ})$.We know that $	an 	heta = \cot(90^{\circ} - 	heta)$.Substituting this into the equation,we

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$36$

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(D) Given the equation: $	an 2A = \cot(A - 18^{\circ})$.We know that $	an 	heta = \cot(90^{\circ} - 	heta)$.Substituting this into the equation,we get: $\cot(90^{\circ} - 2A) = \cot(A - 18^{\circ})$.Since the cotangent functions are equal,their angles must be equal: $90^{\circ} - 2A = A - 18^{\circ}$.Rearranging the terms to solve for $A$: $90^{\circ} + 18^{\circ} = A + 2A$.$108^{\circ} = 3A$.$A = \frac{108^{\circ}}{3} = 36^{\circ}$.Thus,the value of $A$ is $36^{\circ}$.

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

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