+0

# Finding θ When Given Sine

0
300
2

The answer is 45°, that means x has to be $${\sqrt{2}}$$. What did I do wrong?

Jun 11, 2018

#1
+44
+1

Let us say everything is positive. Now the ratio of a 45-45-90 triangle is 1:1:$$\sqrt2$$ . But the ratio on the question is  $$\frac{\sqrt3}{2}$$

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Jun 12, 2018
#2
+7477
+1

Let $$\phi$$ be the reference angle of $$\theta$$.

$$\sin \theta = \dfrac{\sqrt3}{2}\\ \therefore \sin \phi = \dfrac{\sqrt3}{2}\\ \text{As }\phi\text{ is a reference angle, it always lies in Quadrant I.}\\ \therefore \phi = 60^{\circ}\\ \theta\text{ lies in Quadrant IV}\\ \therefore \theta = 360^{\circ}-60^{\circ} = 300^{\circ}\\$$

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Jun 12, 2018