Question

If root 3 sin theta-cos theta=0 and 0 < theta <90°, find the value of theta.

Answer 1

$\bold{\theta=30^{\circ}}$ is the value of $\bold{\sqrt{3} \sin \theta-\cos \theta=0}$ if $\bold{0<\theta<90^{\circ}.}$

Given:

$\sqrt{3} \sin \theta-\cos \theta=0$

$0<\theta<90^{\circ}.$

To find:

Value of θ =?

Solution:

The question is $\sqrt{3} \sin \theta-\cos \theta=0$

Now to solve the question we transfer the cos θ on the other side of the equal to with which we get

$\sqrt{3} \sin \theta-\cos \theta=0$

$\sqrt{3}=\frac{\cos \theta}{\sin \theta}$

$\sqrt{3}=\cot \theta$

Therefore, transferring the cot on the other side of the equal to we get the inverse value of the cot i.e.

$\cot ^{-1} \sqrt{3}=\theta$

Hence, the value of θ is proved to be 30, now the question says the θ is between zero degree and 90 degree thereby, proving that the value of $\bold{\theta=30^{\circ}.}$

Answer 2

Answer:

30°

Step-by-step explanation:

√3sinθ - cosθ = 0

√3sinθ = cosθ

√3 = cosθ / sinθ

√3 = cotθ

but we know that, cot30° = √3

∴ cotθ = cot30°

θ = 30°