Question

consider a triangle ( abc ) like the one below. suppose that ( b = 71 ), ( c = 50 ), and ( b = 29^{circ} ). (the figure is not drawn to scale.) solve the triangle.
carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.
if no such triangle exists, enter
o solution.\ if there is more than one solution, use the button labeled \or\.

Explanation:

Step1: Use the Law of Sines to find angle \(C\)

The Law of Sines states that \(\frac{\sin B}{b}=\frac{\sin C}{c}\).
Substitute \(b = 71\), \(c = 50\), and \(B=29^{\circ}\) into the formula: \(\sin C=\frac{c\sin B}{b}\)
\(\sin C=\frac{50\sin29^{\circ}}{71}\)
\(\sin29^{\circ}\approx0.4848\), so \(\sin C=\frac{50\times0.4848}{71}\approx\frac{24.24}{71}\approx0.3414\)
\(C=\sin^{- 1}(0.3414)\approx20.0^{\circ}\) or \(C = 180^{\circ}-20.0^{\circ}=160.0^{\circ}\)
But if \(C = 160^{\circ}\), then \(B + C=29^{\circ}+160^{\circ}=189^{\circ}>180^{\circ}\), so we reject \(C = 160^{\circ}\)

Step2: Find angle \(A\)

Since the sum of angles in a triangle is \(A + B + C=180^{\circ}\), then \(A=180^{\circ}-B - C\)
Substitute \(B = 29^{\circ}\) and \(C = 20.0^{\circ}\), \(A=180^{\circ}-29^{\circ}-20.0^{\circ}=131.0^{\circ}\)

Step3: Use the Law of Sines to find side \(a\)

Again, by the Law of Sines \(\frac{\sin B}{b}=\frac{\sin A}{a}\)
\(a=\frac{b\sin A}{\sin B}\)
\(\sin A=\sin131^{\circ}\approx0.7547\), \(\sin B=\sin29^{\circ}\approx0.4848\), \(b = 71\)
\(a=\frac{71\times0.7547}{0.4848}\approx\frac{53.5837}{0.4848}\approx110.5\)

Answer:

\(C = 20.0^{\circ},A = 131.0^{\circ},a = 110.5\)