Question

show all work. no work, no credit!
using the given information to determine if a triangle is exists. if one does, then find all remaining values. be sure to als
check for a second triangle when conditions permit. write neatly and circle all answers.

1. ( b = 75 ^ { circ }, c = 37 ^ { circ }, a = 96 )

Explanation:

Step1: Find angle \(A\)

The sum of angles in a triangle is \(A + B + C=180^{\circ}\). Given \(B = 75^{\circ}\) and \(C = 37^{\circ}\), then \(A=180^{\circ}-(B + C)=180^{\circ}-(75^{\circ}+37^{\circ}) = 68^{\circ}\)

Step2: Use the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

To find \(b\):
Since \(\frac{a}{\sin A}=\frac{b}{\sin B}\), and \(a = 96\), \(A = 68^{\circ}\), \(B=75^{\circ}\)
\(b=\frac{a\sin B}{\sin A}=\frac{96\times\sin75^{\circ}}{\sin68^{\circ}}\)
We know that \(\sin75^{\circ}=\sin(45^{\circ}+30^{\circ})=\sin45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659\), \(\sin68^{\circ}\approx0.9272\)
\(b=\frac{96\times0.9659}{0.9272}\approx100\)
To find \(c\):
Since \(\frac{a}{\sin A}=\frac{c}{\sin C}\), and \(C = 37^{\circ}\)
\(c=\frac{a\sin C}{\sin A}=\frac{96\times\sin37^{\circ}}{\sin68^{\circ}}\)
We know that \(\sin37^{\circ}\approx0.6018\), \(\sin68^{\circ}\approx0.9272\)
\(c=\frac{96\times0.6018}{0.9272}\approx62\)

Since \(A + B + C=180^{\circ}\) and we have used the Law of Sines (no ambiguous - case as we are given two angles and one side (ASA - like situation)), there is only one triangle.

Answer:

\(A = 68^{\circ}\), \(b\approx100\), \(c\approx62\)