Ответ:

1) \(c=\sqrt{a^2+b^2}=\sqrt{4^2+3^2}=5\) м.

\(\sin\alpha=\frac{a}{c}=\frac45\), сондықтан \(\alpha\approx53.13^\circ\), \(\beta=90^\circ-\alpha\approx36.87^\circ\).

2) \(b=\sqrt{c^2-a^2}=\sqrt{13^2-12^2}=5\) см.

\(\sin a=\frac{12}{13}\), сондықтан \(\alpha\approx67.38^\circ\), \(\beta\approx22.62^\circ\).

3) \(a=c\sin30^\circ=12\cdot\frac12=6\) см.

\(b=c\cos30^\circ=12\cdot\frac{\sqrt3}{2}=6\sqrt3\) см, \(\beta=60^\circ\).

4) \(\tan60^\circ=\frac{a}{b}\), сондықтан \(a=7\sqrt3\) дм.

\(c=\frac{b}{\cos60^\circ}=14\) дм, \(\beta=30^\circ\).

Жауап: 1) \(c=5\) м, \(\alpha\approx53.13^\circ\), \(\beta\approx36.87^\circ\); 2) \(b=5\) см, \(\alpha\approx67.38^\circ\), \(\beta\approx22.62^\circ\); 3) \(a=6\) см, \(b=6\sqrt3\) см, \(\beta=60^\circ\); 4) \(a=7\sqrt3\) дм, \(c=14\) дм, \(\beta=30^\circ\).