Question

what is the approximate value of b, rounded to the nearest tenth? use the law of sines to find the answer. law of sines: \\( \frac{\sin(a)}{a}=\frac{\sin(b)}{b}=\frac{\sin(c)}{c} \\)

Explanation:

Step1: Find angle \( C \)

In a triangle, \( A + B + C=180^{\circ} \). Given \( A = 66^{\circ}\), \( B = 76^{\circ}\), then \( C=180-(66 + 76)=38^{\circ}\)

Step2: Apply the Law of Sines

The Law of Sines is \(\frac{\sin(A)}{a}=\frac{\sin(B)}{b}=\frac{\sin(C)}{c}\). Here \(a = 3\), \(A = 66^{\circ}\), \(C = 38^{\circ}\), and we want to find \(b\). Using \(\frac{\sin(A)}{a}=\frac{\sin(C)}{b}\), we can rewrite it as \(b=\frac{a\sin(C)}{\sin(A)}\)
Substitute \(a = 3\), \(\sin(A)=\sin(66^{\circ})\approx0.9135\), \(\sin(C)=\sin(38^{\circ})\approx0.6157\)
\(b=\frac{3\times0.6157}{0.9135}=\frac{1.8471}{0.9135}\approx2.02\) (This is wrong approach, correct is using \(\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\))
Correct Step2:
Using \(\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\), where \(a = 3\), \(A = 66^{\circ}\), \(B = 76^{\circ}\)
\(\sin(A)=\sin(66^{\circ})\approx0.9135\), \(\sin(B)=\sin(76^{\circ})\approx0.9703\)
\(b=\frac{3\times\sin(76^{\circ})}{\sin(66^{\circ})}=\frac{3\times0.9703}{0.9135}=\frac{2.9109}{0.9135}\approx3.2\) (wrong again, correct formula \(\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\) with \(a\) opposite \(A\), \(b\) opposite \(B\) no, wait \(a\) is side \(BC\)? No, in standard notation \(a\) is side opposite \(A\), \(b\) is side opposite \(B\). Wait no, in the problem, assume side \(a\) is \(BC\), no, wait in the triangle, side opposite \(A\) is \(BC\) (length \(a\)), side opposite \(B\) is \(AC\) (length \(b\)), side opposite \(C\) is \(AB\) (length \(c = 3\))
So using \(\frac{\sin(A)}{a}=\frac{\sin(C)}{c}\) is wrong. Correct is \(\frac{\sin(A)}{a}=\frac{\sin(B)}{b}\) where \(a\) is side opposite \(A\) (assume \(a\) is \(BC\), no, wait in the problem, \(AB = 3\), \(AB\) is opposite \(C\), \(AC=b\) is opposite \(B\), \(BC\) is opposite \(A\)
Using \(\frac{\sin(C)}{AB}=\frac{\sin(B)}{AC}\) (since \(AB = 3\), \(AC = b\), \(C = 38^{\circ}\), \(B=76^{\circ}\))
\(\frac{\sin(38^{\circ})}{3}=\frac{\sin(76^{\circ})}{b}\)
\(b=\frac{3\sin(76^{\circ})}{\sin(38^{\circ})}\)
\(\sin(76^{\circ})\approx0.9703\), \(\sin(38^{\circ})\approx0.6157\)
\(b=\frac{3\times0.9703}{0.6157}=\frac{2.9109}{0.6157}\approx4.7\)

Answer:

\(4.7\) units