Question

consider a triangle ( abc ) like the one below. suppose that ( b = 25^{circ}, c = 106^{circ} ), and ( a = 48 ). (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\. ( a=square^{circ}, b=square, c=square )

Explanation:

Step1: Find angle \(A\)

The sum of angles in a triangle is \(180^{\circ}\). So \(A = 180^{\circ}-(B + C)\).
Substitute \(B = 25^{\circ}\) and \(C=106^{\circ}\):
\(A=180^{\circ}-(25^{\circ}+ 106^{\circ})=49^{\circ}\)

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

The Law of Sines states \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
We know \(a = 48\), \(A = 49^{\circ}\), \(B = 25^{\circ}\).
So \(b=\frac{a\sin B}{\sin A}\).
\(\sin A=\sin49^{\circ}\approx0.755\), \(\sin B=\sin25^{\circ}\approx0.423\)
\(b=\frac{48\times0.423}{0.755}=\frac{20.304}{0.755}\approx26.9\)

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

By the Law of Sines \(\frac{a}{\sin A}=\frac{c}{\sin C}\).
\(\sin C=\sin106^{\circ}\approx0.970\)
\(c=\frac{a\sin C}{\sin A}=\frac{48\times0.970}{0.755}=\frac{46.56}{0.755}\approx61.7\)

Answer:

\(A = 49^{\circ}\), \(b\approx26.9\), \(c\approx61.7\)