If \(\tan\theta = 5/12\) and \(\theta\) is in Quadrant III, what is the value of \(\sin\theta\)?
\(-5/13\)
We are told \(\tan\theta=\dfrac{5}{12}\). By definition tangent is the ratio of the side opposite the angle to the side adjacent to it, so we can read off opposite \(=5\) and adjacent \(=12\) for a reference right triangle.
The hypotenuse comes from the Pythagoras theorem: \(\text{hyp}=\sqrt{5^{2}+12^{2}}=\sqrt{25+144}=\sqrt{169}=13\). So we are dealing with the well-known 5-12-13 triple.
Sine is opposite over hypotenuse, hence the bare magnitude is \(|\sin\theta|=\dfrac{5}{13}\). This fixes the number; only the sign is still to be decided.
The sign is set by the quadrant. Using the ASTC (All-Silver-Tea-Cups) rule, in Quadrant I all ratios are positive, in Quadrant II only sine, in Quadrant III only tangent, and in Quadrant IV only cosine.
Here \(\theta\) lies in Quadrant III, where only tangent is positive; both sine and cosine are negative. That is consistent with the given positive value of \(\tan\theta\).
Attaching the negative sign to the magnitude gives \(\sin\theta=-\dfrac{5}{13}\).
Key concept: build the numerical ratio from a right triangle (or from \(1+\tan^{2}\theta=\sec^{2}\theta\)), then attach the correct sign using the quadrant through the ASTC rule.
Hence, \(\sin\theta=\) \(-\dfrac{5}{13}\).
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