Question

If tan2A=cot(A−18), where 2A is an acute angle, then find the value of A​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: 2A \: is \: an \: acute \: angle$

and

$\rm \: tan2A = cot(A - 18\degree )$

We know,

$\boxed{ \rm{ \:tanx = cot(90\degree - x) \: }}$

So, using this result, the above expression can be rewritten as

$\rm \: cot(90\degree - 2A) = cot(A - 18\degree )$

On comparing, we get

$\rm\implies \:90\degree - 2A= A - 18\degree$

$\rm \: 2A + A = 90\degree + 18\degree$

$\rm \: 3A = 108\degree$

$\rm\implies \:A = 36\degree$

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Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \\ \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

$\large\underline{\sf{A=36°}}$

Step-by-step explanation:

$\large\underline{\sf{Given \: that}}$

$\large\underline{\sf{2A \: is \: an \: acute \: angle}}$

$\large\underline{\sf{and}}$

$\large\underline{\sf{ \tan(2A) = \cot( A - 18 \degree)}}$

$\large\underline{\sf{We \: know}}$

$\large\boxed{\sf{ \tan x = \cot(90 \degree - x) }}$

So,using this , the above expression can be written as

$\large\underline{\sf{ \cot(90 \degree - 2A) = \cot(A-18°) }}$

$\large\underline{\sf{ ➡\: On \: comparing \: we \: get}}$

$\large\underline{\sf{90°-2A =A - 18 \degree }}$

$\large\underline{\sf{ \implies 3A = 108 \degree}}$

$\large\underline{\sf{A = 36 \degree}}$