Using \(\bigtriangleup\)XYZ in the figure, find XYZ
29o
31o
31o 20'
31o 18'
Correct answer is C
\(\frac{\sin y}{3} = \frac{\sin 120^o}{5}\)
sin 120o = sin 60o
5 sin y = 3 sin 60o
sin y = \(\frac{3 \sin 60^o}{5}\)
\(\frac{3 \times 0.866}{5}\)
= \(\frac{2.598}{5}\)
y = sin-1 0.5196 = 30o 18'
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