Solution
$$\sin \theta = \cos(50^\circ + \theta)$$
$$\cos\left(90^{\circ\ }-\theta\ \right)=\cos\left(50^{\circ\ }+\theta\ \right)$$
$$90^{\circ\ }-\theta\ =50^{\circ\ }+\theta\ $$
$$2\theta\ =40^{\circ\ }$$
$$\theta\ =20^{\circ\ }$$
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