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If √3 sinteta − costeta=0 and 0 < ߠ > 90eta, find the value of teta. |
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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∘ . Step-by-step EXPLANATION: .. thanku.. Mark as brainliest |
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