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.

Explanation:

Step1: Find angle Q

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

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{o}{\sin O}=\frac{q}{\sin Q}\). We know \(q = 75\) cm, \(m\angle O = 113^\circ\), \(m\angle Q = 49^\circ\). So we can solve for \(o\):
\(o=\frac{q\times\sin O}{\sin Q}\)
Substitute the values: \(\sin 113^\circ\approx\sin(90^\circ + 23^\circ)=\cos 23^\circ\approx0.9205\), \(\sin 49^\circ\approx0.7547\), \(q = 75\)
\(o=\frac{75\times0.9205}{0.7547}\)

Step3: Calculate the value of o

First, calculate the numerator: \(75\times0.9205 = 69.0375\)
Then divide by the denominator: \(o=\frac{69.0375}{0.7547}\approx91.48\)
Rounding to the nearest centimeter, \(o\approx91\)

Answer:

\(91\)