Question

two sides and an angle are given. determine whether the given information results in one triangle, two triangles, or no triangle at all. solve any resulting triangle(s)
a = 10, b = 9, a = 20°

Explanation:

Step1: Use the Law of Sines

By the Law of Sines, \(\frac{\sin B}{b}=\frac{\sin A}{a}\). Substitute \(a = 10\), \(b = 9\), \(A=20^{\circ}\). So \(\sin B=\frac{b\sin A}{a}=\frac{9\sin20^{\circ}}{10}\).
Calculate \(\sin20^{\circ}\approx0.3420\), then \(\sin B=\frac{9\times0.3420}{10}=0.3078\).

Step2: Find angle \(B\)

Since \(\sin B = 0.3078\), then \(B=\sin^{-1}(0.3078)\approx18^{\circ}\) or \(B = 180^{\circ}- 18^{\circ}=162^{\circ}\). But \(A = 20^{\circ}\), if \(B = 162^{\circ}\), then \(A + B=20^{\circ}+162^{\circ}=182^{\circ}>180^{\circ}\). So we discard \(B = 162^{\circ}\).

Step3: Find angle \(C\)

Since \(A + B + C=180^{\circ}\), \(A = 20^{\circ}\), \(B\approx18^{\circ}\), then \(C=180^{\circ}-(20^{\circ}+18^{\circ})=142^{\circ}\).

Step4: Use the Law of Sines to find \(c\)

By the Law of Sines \(\frac{c}{\sin C}=\frac{a}{\sin A}\). So \(c=\frac{a\sin C}{\sin A}\). Substitute \(a = 10\), \(A = 20^{\circ}\), \(C = 142^{\circ}\), \(\sin142^{\circ}=\sin(180^{\circ}-38^{\circ})=\sin38^{\circ}\approx0.6157\), \(\sin20^{\circ}\approx0.3420\). Then \(c=\frac{10\times0.6157}{0.3420}\approx18\).

Answer:

One triangle. \(B\approx18^{\circ}\), \(C = 142^{\circ}\), \(c\approx18\)