Question 73

If $$\sin \theta = \cos(50^\circ + \theta)$$, then $$\theta$$ is equal to:

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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