Question

consider a triangle abc like the one below. suppose that a = 51, b = 66, and a = 32°. (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 \( B \)

The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
Substituting the given values \(a = 51\), \(b = 66\), and \(A=32^{\circ}\), we get \(\sin B=\frac{b\sin A}{a}\).
\(\sin B=\frac{66\sin32^{\circ}}{51}\)
\(\sin B=\frac{66\times0.5299}{51}\)
\(\sin B=\frac{34.9734}{51}\approx0.6857\)
\(B=\sin^{- 1}(0.6857)\approx43.3^{\circ}\) or \(B = 180^{\circ}-43.3^{\circ}=136.7^{\circ}\)

Step2: Check for valid triangles

If \(B = 43.3^{\circ}\):

• Find angle \(C\): \(C=180^{\circ}-A - B=180^{\circ}-32^{\circ}-43.3^{\circ}=104.7^{\circ}\)
• Use the Law of Sines to find \(c\): \(\frac{c}{\sin C}=\frac{a}{\sin A}\), so \(c=\frac{a\sin C}{\sin A}\)

\(c=\frac{51\sin104.7^{\circ}}{\sin32^{\circ}}=\frac{51\times0.9703}{0.5299}=\frac{49.4853}{0.5299}\approx93.4\)

If \(B = 136.7^{\circ}\):

• Find angle \(C\): \(C=180^{\circ}-A - B=180^{\circ}-32^{\circ}-136.7^{\circ}=11.3^{\circ}\)
• Use the Law of Sines to find \(c\): \(\frac{c}{\sin C}=\frac{a}{\sin A}\), so \(c=\frac{a\sin C}{\sin A}\)

\(c=\frac{51\sin11.3^{\circ}}{\sin32^{\circ}}=\frac{51\times0.1951}{0.5299}=\frac{9.9501}{0.5299}\approx18.8\)

Answer:

\(B = 43.3^{\circ},C = 104.7^{\circ},c = 93.4\) or \(B = 136.7^{\circ},C = 11.3^{\circ},c = 18.8\)