Question

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If tan 2A = cot (A - 18°), Where 2A is acute, Find the value of A

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Answer 1

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Given $tan2A = cot(A−18⁰)$

$⇒ cot(90 - 2 A) = cot( A - {18}^{0} )$

$[∵tanθ=cot(90−θ)]$

Comparing Angles we get :-

$90 - 2A = A - 18$

$⇒90 + 18 = 2A + A$

$⇒3A = 108$

$⇒A = \frac{108}{3}$

$⇒A = {36}^{o}$

Answer 2

Answer:

Given tan2A = cot(A−18⁰)tan2A=cot(A−18⁰)

⇒ cot(90 - 2 A) = cot( A - {18}^{0} )⇒cot(90−2A)=cot(A−18

0

)

[∵tanθ=cot(90−θ)][∵tanθ=cot(90−θ)]

Comparing Angles we get :-

90 - 2A = A - 1890−2A=A−18

⇒90 + 18 = 2A + A⇒90+18=2A+A

⇒3A = 108⇒3A=108

⇒A = \frac{108}{3}⇒A=

3

108

⇒A = {36}^{o}⇒A=36

o

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