If sin 5 theta = cos 20^{circ} where 0^{circ} | vedclass

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(D) Given the equation: $\sin 5 	heta = \cos 20^{\circ}$.We know the trigonometric identity: $\cos A = \sin(90^{\circ} - A)$.Applying this identity t

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$14$

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(D) Given the equation: $\sin 5 	heta = \cos 20^{\circ}$.We know the trigonometric identity: $\cos A = \sin(90^{\circ} - A)$.Applying this identity to the right side: $\cos 20^{\circ} = \sin(90^{\circ} - 20^{\circ}) = \sin 70^{\circ}$.Now,the equation becomes: $\sin 5 	heta = \sin 70^{\circ}$.By comparing the angles,we get: $5 	heta = 70^{\circ}$.Solving for $	heta$: $	heta = \frac{70^{\circ}}{5} = 14^{\circ}$.Thus,the value of $	heta$ is $14^{\circ}$.

If $\sin 5 \theta = \cos 20^{\circ}$ where $0^{\circ} < \theta < 90^{\circ}$,then the value of $\theta$ is (in degrees):

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