Question - Solving for the Sine of Angle R in a Right Triangle Given the Tangent of Angle S
Solution:
Given that $$\tan(S) = \frac{1}{\sqrt{3}}$$ and $$\angle T = 90$$ degrees, so $$\angle S + \angle R = 90$$ degrees.$$\tan(S) = \frac{1}{\sqrt{3}} = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sin(S)}{\cos(S)}$$$$\sin(S) = \cos(R)$$ since $$\sin(\theta) = \cos(90 - \theta)$$Therefore, $$\sin(S) = \frac{1}{\sqrt{3}}$$Now, since $$\sin(S) = \frac{1}{\sqrt{3}}$$, we have $$\sin(R) = \sin(90 - S)$$$$\sin(R) = \cos(S) = \sqrt{1 - \sin^2(S)}$$Calculating $$\cos(S)$$ given $$\sin(S)$$:$$\cos(S) = \sqrt{1 - \left(\frac{1}{\sqrt{3}}\right)^2}$$$$\cos(S) = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}}$$$$\sin(R) = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}$$$$\sin(R) = \frac{\sqrt{6}}{3}$$Since this option is not provided in the multiple-choice answers, it must be simplified further:$$\sin(R) = \frac{\sqrt{6}}{3} = \frac{2 \cdot \sqrt{6}}{2 \cdot 3}$$$$\sin(R) = \frac{2\sqrt{6}}{6}$$$$\sin(R) = \frac{\sqrt{6}}{3}$$Hence, the correct answer is $$\sin(R) = \frac{\sqrt{6}}{3}$$, which is not listed in the provided options, there might be an error in the provided options or question setup.
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