Question

if A and 3A-30 are acute angles such that sinA=cos(3A-30), find the value of tanA

Answer 1

solution:

sin A= cos(3A-30) [Given]

=> sin A= sin [90°-(3A-30)]

=> A= 120 - 3A

=> 4A= 120°

=> A= 30°

Now, tan A [To find]

= tan 30°

$= \frac{1}{ \sqrt{3} }$
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Answer 2

The value of $\tan A$ is $\dfrac{1}{\sqrt{3} }.$

Step-by-step explanation:

We have,

$\sin A=\cos (3A-30)$

To find, the value of $\tan A=?$

$\sin A=\sin [90-(3A-30)]$

[ ∵ $\sin A=\cos (90-A)$]

⇒ $A=90-(3A-30)$

⇒ $A=90-3A+30=120-3A$

⇒ $A+3A=120$

⇒ A = $\dfrac{120}{4}$=30°

∴ $\tan A=\tan 30=\dfrac{1}{\sqrt{3} }$

Hence,  the value of $\tan A$ is $\dfrac{1}{\sqrt{3} }.$