If sin 70^{circ} = cos theta,then theta = ldo | vedclass

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(C) We know the trigonometric identity for complementary angles: $\sin(90^{\circ} - A) = \cos A$ and $\cos(90^{\circ} - A) = \sin A$.Given the equatio

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$20$

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(C) We know the trigonometric identity for complementary angles: $\sin(90^{\circ} - A) = \cos A$ and $\cos(90^{\circ} - A) = \sin A$.Given the equation: $\sin 70^{\circ} = \cos 	heta$.We can rewrite $\sin 70^{\circ}$ as $\cos(90^{\circ} - 70^{\circ})$.Therefore,$\cos(90^{\circ} - 70^{\circ}) = \cos 	heta$.This simplifies to $\cos 20^{\circ} = \cos 	heta$.By comparing the angles,we get $	heta = 20^{\circ}$.

$1. \cos \theta$	$a. \frac{\cos \theta}{\sin \theta}$
$2. \tan \theta$	$b. \frac{1}{\csc \theta}$
$3. \cot \theta$	$c. \frac{1}{\sec \theta}$
$4. \sin \theta$	$d. \frac{1}{\cot \theta}$
	$e. \sin \theta \cdot \cos \theta$

If $\sin 70^{\circ} = \cos \theta$,then $\theta = \ldots \ldots \ldots \ldots$ (in $^{\circ}$)

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