Answer: [tex]AB=4.7, BC=8.8, \angle C=28^{\circ}[/tex]
Step-by-step explanation:
As angles in a triangle add to 180 degrees,
[tex]\angle C=180^{\circ}-90^{\circ}-62^{\circ}=\boxed{28^{\circ}}[/tex]
We know that:
[tex]\sin 62^{\circ}=\frac{BC}{10}\\\\BC=10\sin 62^{\circ} \approx \boxed{8.8}[/tex]
Similarly,
[tex]\cos 62^{\circ}=\frac{AB}{10}\\\\AB=10\cos 62^{\circ} \approx \boxed{4.7}[/tex]
