Question : If $\cot 3 \mathrm{~A}=\tan \left(\mathrm{A}-36^{\circ}\right)$, then what is the value of $\mathrm{A}$?
Option 1: 33.5 degrees
Option 2: 25 degrees
Option 3: 30 degrees
Option 4: 31.5 degrees
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Correct Answer: 31.5 degrees
Solution :
$\cot 3 \mathrm{~A}=\tan \left(\mathrm{A}-36^{\circ}\right)$
$⇒ \tan (90-3 \mathrm{~A})=\tan \left(\mathrm{A}-36^{\circ}\right)$
$⇒ 90^{\circ}-3 \mathrm{~A} = \mathrm{~A}-36^{\circ}$
$⇒ 4 \mathrm{~A} = 126^{\circ}$
$\therefore \mathrm{~A} = 31.5^{\circ}$
Hence, the correct answer is 31.5 degrees.
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