Question

consider a triangle abc like the one below. suppose that a=60°, b=22, and c=56. (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 there is more than one solution, use the button labeled \or\.

Explanation:

Step1: Find angle \( B \)

We know that the sum of angles in a triangle is \( 180^{\circ} \). So \( B=180^{\circ}-A - C \). Given \( A = 60^{\circ}\) and \( C=100^{\circ}\), then \( B=180^{\circ}-60^{\circ}-100^{\circ}=20^{\circ}\)

Step2: Use the Law of Sines to find side \( a \)

The Law of Sines is \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). We want to find \( a \), and we know \( b = 22\), \(A = 60^{\circ}\), \(B=20^{\circ}\). From \(\frac{a}{\sin A}=\frac{b}{\sin B}\), we can solve for \( a\). So \( a=\frac{b\sin A}{\sin B}\).
Substitute \( b = 22\), \(A = 60^{\circ}\) (\(\sin A=\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.8660\)), \(B = 20^{\circ}\) (\(\sin B=\sin20^{\circ}\approx0.3420\)) into the formula:
\( a=\frac{22\times\sin60^{\circ}}{\sin20^{\circ}}=\frac{22\times0.8660}{0.3420}\approx56.0\)

Answer:

\( B = 20^{\circ}\), \(C = 100^{\circ}\), \(a\approx56.0\)