Question

in \\( \triangle efg, e = 75 \mathrm { cm }, m \angle g = 141 ^ { \circ } \\) and \\( m \angle e = 16 ^ { \circ } \\). find the length of \\( g \\), to the nearest centimeter.

Explanation:

Step1: Find the measure of angle \(F\)

In a triangle, the sum of angles is \(180^{\circ}\). So, \(m\angle F=180^{\circ}-(m\angle E + m\angle G)\).
Substituting the given values: \(m\angle F = 180^{\circ}-(16^{\circ}+141^{\circ})=180^{\circ}-157^{\circ} = 23^{\circ}\).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{e}{\sin E}=\frac{g}{\sin G}\).
We know \(e = 75\) cm, \(m\angle E=16^{\circ}\), \(m\angle G = 141^{\circ}\).
From \(\frac{e}{\sin E}=\frac{g}{\sin G}\), we can solve for \(g\): \(g=\frac{e\times\sin G}{\sin E}\).
Substitute the values: \(\sin16^{\circ}\approx0.2756\), \(\sin141^{\circ}=\sin(180 - 39)^{\circ}=\sin39^{\circ}\approx0.6293\), \(e = 75\).
\(g=\frac{75\times0.6293}{0.2756}\).
Calculate \(75\times0.6293 = 47.1975\).
Then \(g=\frac{47.1975}{0.2756}\approx171\).

Answer:

\(171\)