(C) Given that $\operatorname{cosec} \theta = \frac{2}{\sqrt{3}}$.Since $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$,we have $\sin \theta = \

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(C) Given that $\operatorname{cosec} \theta = \frac{2}{\sqrt{3}}$.Since $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$,we have $\sin \theta = \frac{1}{\operatorname{cosec} \theta} = \frac{\sqrt{3}}{2}$.We know from trigonometric standard values that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.Comparing both sides,we get $\theta = 60^\circ$.

Class 10 Mathematics Introduction to Trigonometry Mix Examples - Introduction to Trigonometry

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If $\operatorname{cosec} \theta = \frac{2}{\sqrt{3}}$,then $\theta = \ldots$ (in $^\circ$)

A
$30$
B
$45$
C
$60$
D
$90$

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Introduction to Trigonometry

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Mix Examples - Introduction to Trigonometry

In $\Delta ABC$, $AC = 5$, $BC = 13$, $m \angle A = 90^\circ$, then $\tan B = \ldots$

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For $0 < \theta < 90^{\circ},$ the value of $\ldots \ldots \ldots \ldots$ increases as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.

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$\frac{1}{\cos ^{2} \theta}-1 = \ldots$

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If $\operatorname{cosec} A = \sqrt{10}$,then $\sin A = \dots$

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Prove that,
$\frac{\tan A}{1+\sec A} + \frac{\tan A}{\sec A-1} = 2 \operatorname{cosec} A$

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If $\operatorname{cosec} \theta = \frac{2}{\sqrt{3}}$,then $\theta = \ldots$ (in $^\circ$)

Similar Questions

In $\Delta ABC$, $AC = 5$, $BC = 13$, $m \angle A = 90^\circ$, then $\tan B = \ldots$

For $0 < \theta < 90^{\circ},$ the value of $\ldots \ldots \ldots \ldots$ increases as $\theta$ increases from $0^{\circ}$ to $90^{\circ}$.

$\frac{1}{\cos ^{2} \theta}-1 = \ldots$

If $\operatorname{cosec} A = \sqrt{10}$,then $\sin A = \dots$

Prove that,$\frac{\tan A}{1+\sec A} + \frac{\tan A}{\sec A-1} = 2 \operatorname{cosec} A$

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Prove that,
$\frac{\tan A}{1+\sec A} + \frac{\tan A}{\sec A-1} = 2 \operatorname{cosec} A$