Question

consider a triangle abc like the one below. suppose that a = 76°, c = 39°, and c = 54. (the figure is not drawn to scale.) solve the triangle. round your answers to the nearest tenth. if there is more than one solution, use the button labeled \or\.

Explanation:

Step1: Find angle B

The sum of angles in a triangle is \(180^\circ\). So, \(B = 180^\circ - A - C\).
Given \(A = 76^\circ\) and \(C = 39^\circ\), then \(B = 180 - 76 - 39 = 65^\circ\).

Step2: Use the Law of Sines to find side a

The Law of Sines states that \(\frac{a}{\sin A}=\frac{c}{\sin C}\).
We know \(c = 54\), \(A = 76^\circ\), \(C = 39^\circ\).
So, \(a=\frac{c\sin A}{\sin C}=\frac{54\sin 76^\circ}{\sin 39^\circ}\).
Calculate \(\sin 76^\circ\approx0.9703\), \(\sin 39^\circ\approx0.6293\).
Then \(a=\frac{54\times0.9703}{0.6293}\approx\frac{52.3962}{0.6293}\approx83.3\).

Step3: Use the Law of Sines to find side b

Using the Law of Sines \(\frac{b}{\sin B}=\frac{c}{\sin C}\).
We know \(B = 65^\circ\), \(c = 54\), \(C = 39^\circ\).
So, \(b=\frac{c\sin B}{\sin C}=\frac{54\sin 65^\circ}{\sin 39^\circ}\).
Calculate \(\sin 65^\circ\approx0.9063\).
Then \(b=\frac{54\times0.9063}{0.6293}\approx\frac{48.9402}{0.6293}\approx77.8\).

Answer:

\(B = 65^\circ\), \(a\approx83.3\), \(b\approx77.8\)