C-09-Q197mediumsingle_mcq

If $\cot(\theta - 30^\circ) = \dfrac{1}{\sqrt{3}}$ , then $\sin\theta = ?$

a $\dfrac{1}{2}$
b $\dfrac{\sqrt{3}}{2}$
c $1$
d $0$

ব্যাখ্যা

From $\cot(\theta-30^\circ)=\tfrac{1}{\sqrt3}$ we get $\tan(\theta-30^\circ)=\sqrt3$ , so $\theta-30^\circ=60^\circ$ and $\theta=90^\circ$ . Hence $\sin\theta=\sin 90^\circ=1$ .

এই অধ্যায়ের আরও প্রশ্ন

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Trigonometric Ratio — সব প্রশ্ন দেখুন (217টি) →

If $\cot(\theta - 30^\circ) = \dfrac{1}{\sqrt{3}}$ , then $\sin\theta = ?$

\dfrac{1}{2}

\dfrac{\sqrt{3}}{2}

1

0

ব্যাখ্যা

From $\cot(\theta-30^\circ)=\tfrac{1}{\sqrt3}$ we get $\tan(\theta-30^\circ)=\sqrt3$ , so $\theta-30^\circ=60^\circ$ and $\theta=90^\circ$ . Hence $\sin\theta=\sin 90^\circ=1$ .

If cot⁡(θ−30∘)=13\cot(\theta - 30^\circ) = \dfrac{1}{\sqrt{3}}cot(θ−30∘)=3​1​, then sin⁡θ=?\sin\theta = ?sinθ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

If cot⁡(θ−30∘)=13\cot(\theta - 30^\circ) = \dfrac{1}{\sqrt{3}}cot(θ−30∘)=3​1​, then sin⁡θ=?\sin\theta = ?sinθ=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

If $\cot(\theta - 30^\circ) = \dfrac{1}{\sqrt{3}}$ , then $\sin\theta = ?$

If $\cot(\theta - 30^\circ) = \dfrac{1}{\sqrt{3}}$ , then $\sin\theta = ?$