Question

consider a triangle abc like the one below. suppose that b = 109°, c = 33°, and c = 49. (the figure is not drawn to scale.) solve the triangle. round your answers to the nearest tenth. if there is more than one solution, use the button labeled \or\.

Explanation:

Step1: Find angle \(A\)

Use the angle - sum property of a triangle \(A + B + C=180^{\circ}\).
\(A = 180^{\circ}-(B + C)\)
Substitute \(B = 109^{\circ}\) and \(C = 33^{\circ}\)
\(A=180^{\circ}-(109^{\circ}+ 33^{\circ})=38^{\circ}\)

Step2: Use the Law of Sines \(\frac{a}{\sin A}=\frac{c}{\sin C}\) to find \(a\)

We know that \(C = 33^{\circ}\), \(c = 49\), \(A = 38^{\circ}\)
From \(\frac{a}{\sin A}=\frac{c}{\sin C}\), we can solve for \(a\):
\(a=\frac{c\sin A}{\sin C}\)
\(\sin A=\sin(38^{\circ})\approx0.616\), \(\sin C=\sin(33^{\circ})\approx0.545\)
\(a=\frac{49\times0.616}{0.545}=\frac{30.184}{0.545}\approx55.4\)

Step3: Use the Law of Sines \(\frac{b}{\sin B}=\frac{c}{\sin C}\) to find \(b\)

We know that \(B = 109^{\circ}\), \(c = 49\), \(\sin B=\sin(109^{\circ})\approx0.946\), \(\sin C=\sin(33^{\circ})\approx0.545\)
From \(\frac{b}{\sin B}=\frac{c}{\sin C}\), we can solve for \(b\):
\(b=\frac{c\sin B}{\sin C}\)
\(b=\frac{49\times0.946}{0.545}=\frac{46.354}{0.545}\approx85.1\)

Answer:

\(A = 38^{\circ}\), \(a\approx55.4\), \(b\approx85.1\)