Question

solve the triangle.
(image of triangle abc with angle a = 31°, angle c = 78°, side a = 8)
b = \\(\square\\)°
(do not round until the final answer. then round to the nearest degree as needed.)
b ≈ \\(\square\\)
(do not round until the final answer. then round to the nearest tenth as needed.)
c ≈ \\(\square\\)
(do not round until the final answer. then round to the nearest tenth as needed.)

Explanation:

Step1: Find angle B

The sum of angles in a triangle is \(180^\circ\). So, \(B = 180^\circ - A - C\). Given \(A = 31^\circ\) and \(C = 78^\circ\), we calculate \(B = 180 - 31 - 78 = 71^\circ\).

Step2: Find side b using the Law of Sines

The Law of Sines states \(\frac{a}{\sin A}=\frac{b}{\sin B}\). We know \(a = 8\), \(A = 31^\circ\), and \(B = 71^\circ\). So, \(b=\frac{a\sin B}{\sin A}=\frac{8\sin 71^\circ}{\sin 31^\circ}\). First, calculate \(\sin 71^\circ\approx0.9455\) and \(\sin 31^\circ\approx0.5150\). Then, \(b=\frac{8\times0.9455}{0.5150}=\frac{7.564}{0.5150}\approx14.7\).

Step3: Find side c using the Law of Sines

Using the Law of Sines \(\frac{a}{\sin A}=\frac{c}{\sin C}\). So, \(c=\frac{a\sin C}{\sin A}=\frac{8\sin 78^\circ}{\sin 31^\circ}\). Calculate \(\sin 78^\circ\approx0.9781\). Then, \(c=\frac{8\times0.9781}{0.5150}=\frac{7.8248}{0.5150}\approx15.2\).

Answer:

\(B = 71^\circ\)
\(b\approx14.7\)
\(c\approx15.2\)