Question

question
in \\(\triangle opq\\), \\(q = 75\\) cm, \\(m\angle o = 113^\circ\\), and \\(m\angle p = 18^\circ\\). find the length of \\(o\\), to the nearest centimeter.
answer attempt 1 out of 3
\\(o = \boxed{\space}\\) cm
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Explanation:

Step1: Find angle Q

In a triangle, the sum of angles is \(180^\circ\). So \(m\angle Q = 180^\circ - m\angle O - m\angle P\).
\(m\angle Q = 180 - 113 - 18 = 49^\circ\)

Step2: Apply the Law of Sines

The Law of Sines states \(\frac{o}{\sin O}=\frac{q}{\sin Q}\). We know \(q = 75\) cm, \(m\angle O = 113^\circ\), \(m\angle Q = 49^\circ\).
So \(o=\frac{q\times\sin O}{\sin Q}\)
\(\sin 113^\circ\approx\sin(90^\circ + 23^\circ)=\cos 23^\circ\approx0.9205\)
\(\sin 49^\circ\approx0.7547\)
\(o=\frac{75\times0.9205}{0.7547}\approx\frac{69.0375}{0.7547}\approx91.48\approx91\) (rounded to nearest centimeter)

Answer:

91