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

 Jun 11, 2018
 #1
avatar+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
avatar+7709 
+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

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