Question

1. in \\( \triangle a b c, \angle a = 58 ^ { \circ }, a c = 22 \\), and \\( b c = 25 \\). solve the triangle.

\\( \angle b = \\)
\\( \angle c = \\)
\\( a b = \\)

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that \(\frac{BC}{\sin A}=\frac{AC}{\sin B}\).
Substitute the given values: \(\frac{25}{\sin58^{\circ}}=\frac{22}{\sin B}\).
Then \(\sin B=\frac{22\sin58^{\circ}}{25}\).
Calculate \(\sin58^{\circ}\approx0.848\), so \(\sin B=\frac{22\times0.848}{25}=\frac{18.656}{25} = 0.74624\).
\(B=\sin^{- 1}(0.74624)\approx48.3^{\circ}\) or \(B = 180^{\circ}-48.3^{\circ}=131.7^{\circ}\).
But if \(B = 131.7^{\circ}\), then \(A + B=58^{\circ}+131.7^{\circ}=189.7^{\circ}>180^{\circ}\) (violates the triangle - angle sum property \(A + B + C=180^{\circ}\)). So \(B\approx48.3^{\circ}\).

Step2: Find angle \(C\)

Using the triangle - angle sum property \(A + B + C=180^{\circ}\).
\(C=180^{\circ}-A - B\).
Substitute \(A = 58^{\circ}\) and \(B\approx48.3^{\circ}\), then \(C=180^{\circ}-58^{\circ}-48.3^{\circ}=73.7^{\circ}\).

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

By the Law of Sines \(\frac{AB}{\sin C}=\frac{BC}{\sin A}\).
\(AB=\frac{BC\sin C}{\sin A}\).
Substitute \(BC = 25\), \(\sin C\approx\sin73.7^{\circ}\approx0.96\), \(\sin A=\sin58^{\circ}\approx0.848\).
\(AB=\frac{25\times0.96}{0.848}=\frac{24}{0.848}\approx28.3\).

Answer:

\(\angle B\approx48.3^{\circ}\), \(\angle C\approx73.7^{\circ}\), \(AB\approx28.3\)