Question

If √3 sinteta − costeta=0 and 0 < ߠ > 90eta, find the value of teta.

Answer 1

Answer:

=30

∘

is the value of \bold{\sqrt{3} \sin \theta-\cos \theta=0}

3

sinθ−cosθ=0 if \bold{0 < \theta < 90^{\circ}.}0<θ<90

∘

.

Given:

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

3

sinθ−cosθ=0

0 < \theta < 90^{\circ}.0<θ<90

∘

.

To find:

Value of θ =?

Solution:

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

3

sinθ−cosθ=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

3

sinθ−cosθ=0

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

3

=

sinθ

cosθ

\sqrt{3}=\cot \theta

3

=cotθ

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}=\thetacot

−1

3

=θ

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}.}θ=30

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Step-by-step explanation:

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