Question

consider a triangle abc like the one below. suppose that b=54, c=64, and b=25°. (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\. c = □°, a = □°, a = □

Explanation:

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

The Law of Sines states that \(\frac{\sin B}{b}=\frac{\sin C}{c}\).
We know that \(b = 54\), \(c=64\), and \(B = 25^{\circ}\).
So, \(\sin C=\frac{c\sin B}{b}=\frac{64\sin25^{\circ}}{54}\).
Since \(\sin25^{\circ}\approx0.4226\), then \(\sin C=\frac{64\times0.4226}{54}\approx\frac{27.0464}{54}\approx0.5009\).
\(C=\sin^{- 1}(0.5009)\approx30.0^{\circ}\) or \(C = 180^{\circ}-30.0^{\circ}=150.0^{\circ}\). But if \(C = 150^{\circ}\), then \(B + C=25^{\circ}+150^{\circ}=175^{\circ}\lt180^{\circ}\).

Step2: Find angle \(A\)

Using the angle - sum property of a triangle \(A + B + C=180^{\circ}\).
If \(C = 30.0^{\circ}\), then \(A=180^{\circ}-(25^{\circ}+30^{\circ}) = 125.0^{\circ}\).
If \(C = 150.0^{\circ}\), then \(A=180^{\circ}-(25^{\circ}+150^{\circ})=5.0^{\circ}\).

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

Again, by the Law of Sines \(\frac{\sin A}{a}=\frac{\sin B}{b}\), so \(a=\frac{b\sin A}{\sin B}\).
Case 1: When \(A = 125.0^{\circ}\), \(\sin A=\sin125^{\circ}\approx0.8192\), \(b = 54\), \(\sin B=\sin25^{\circ}\approx0.4226\).
\(a=\frac{54\times0.8192}{0.4226}\approx\frac{44.2368}{0.4226}\approx104.7\).
Case 2: When \(A = 5.0^{\circ}\), \(\sin A=\sin5^{\circ}\approx0.0872\).
\(a=\frac{54\times0.0872}{0.4226}\approx\frac{4.7088}{0.4226}\approx11.1\).

Answer:

\(C = 30.0^{\circ},A = 125.0^{\circ},a = 104.7\) or \(C = 150.0^{\circ},A = 5.0^{\circ},a = 11.1\)