Question

find the length of side a.
b = 58° a = ? c =
round to the nearest tenth.

Explanation:

Step1: Find angle \( B \)

The sum of angles in a triangle is \( 180^{\circ} \). Given \( A = 109^{\circ} \), \( C=13^{\circ} \), then \( B = 180^{\circ}-(109^{\circ}+13^{\circ})=58^{\circ} \) (already given).

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, we want to find \( a \), and we know \( A = 109^{\circ} \), \( C = 13^{\circ} \), and the side opposite \( C \) (let's assume the side of length \( 6 \) is opposite \( B\) is wrong, no, wait, no - wait, no, the side of length \( 6\) is \( AC\), wait no, in standard notation, side \( a\) is opposite \( A\), side \( b\) is opposite \( B\), side \( c\) is opposite \( C\). Wait, no, actually, in the Law of Sines formula \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). If we assume the side adjacent to \( A\) and \( C\) is \( b = 6\) (opposite \( B\)). Then \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Substitute \( A = 109^{\circ}\), \( B = 58^{\circ}\), \( b = 6\). So \(a=\frac{6\times\sin(109^{\circ})}{\sin(58^{\circ})}\).
We know that \(\sin(109^{\circ})=\sin(180 - 71^{\circ})=\sin(71^{\circ})\approx0.946\), \(\sin(58^{\circ})\approx0.848\). Then \(a=\frac{6\times0.946}{0.848}=\frac{5.676}{0.848}\approx6.7\)

Answer:

\(6.7\)