Question

if √3 sinA-cosA=0, find the value of A​

Answer 1

Question

If $\sqrt{3} sin(A)-cos(A)=0$, find the value of A.

Let's divide both sides by $cos(A)$.

According to the identity $tan(A)=\dfrac{sin(A)}{cos(A)}$ we get $\sqrt{3} tan(A)-1=0$.

Therefore $tan(A)=\dfrac{1}{\sqrt{3} }$.

This equation is satisfied if $\angle {A}=30\textdegree$.

The tangent function is $\pi$-periodic, so full answer will be $\pi n+\dfrac{1}{6} \pi$ rad.

More information:

There are three basic trigonometric functions.

Sine, cosine and tangent.

$\implies sin(\theta)=\dfrac{opposite}{hypotenuse}$

$\implies cos(\theta)=\dfrac{adjacent}{hypotenuse}$

$\implies tan(\theta)=\dfrac{opposite}{adjacent}$

If we divide sine by cosine, the denominator of the ratio cancels out.

$\dfrac{sin(\theta)}{cos(\theta)}=\dfrac{opposite}{adjacent}$

This is equal to tangent.

We get a trigonometric identity $\dfrac{sin(\theta)}{cos(\theta)}=tan(\theta)$.

Answer 2

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