Question

1. solve for side b.

Explanation:

Step1: Apply the Law of Cosines

The Law of Cosines states that \(c^{2}=a^{2}+b^{2}-2ab\cos C\). In \(\triangle ABC\), if \(a = 25\), \(c=22\), and \(C = 41^{\circ}\), then \(b^{2}=a^{2}+c^{2}-2ac\cos B\).
Substitute \(a = 25\), \(c = 22\), and \(B=41^{\circ}\) into the formula:
\(b^{2}=25^{2}+22^{2}-2\times25\times22\times\cos41^{\circ}\)

Step2: Calculate each term

• Calculate \(25^{2}=625\), \(22^{2} = 484\)
• Calculate \(2\times25\times22=1100\), and \(\cos41^{\circ}\approx0.7547\)
• Then \(2\times25\times22\times\cos41^{\circ}\approx1100\times0.7547 = 830.17\)
• Now \(b^{2}=625 + 484-830.17\)
• \(b^{2}=1109 - 830.17=278.83\)

Step3: Solve for \(b\)

Take the square root of both sides: \(b=\sqrt{278.83}\approx16.7\) (rounded to one decimal place)

Answer:

\(b\approx16.7\space cm\)