Bài 62:
\[
\sin(C) = \frac{AB}{BC} \implies AB = BC \cdot \sin(30^\circ)
\]
\[
AB = 5 \cdot \frac{1}{2} = 2.5 \, \text{cm}
\]
\[
\cos(C) = \frac{AC}{BC} \implies AC = BC \cdot \cos(30^\circ)
\]
\[
AC = 5 \cdot \frac{\sqrt{3}}{2} \approx 5 \cdot 0.866 = 4.33 \, \text{cm}
\]
Bài 63:
\[
\sin(F) = \frac{DF}{DE} \implies DF = DE \cdot \sin(47^\circ)
\]
\[
DF = 9 \cdot \sin(47^\circ) \approx 9 \cdot 0.731 = 6.58 \, \text{cm}
\]
\[
\cos(F) = \frac{DF}{DE} \implies EF = DE \cdot \cos(47^\circ)
\]
\[
EF = 9 \cdot \cos(47^\circ) \approx 9 \cdot 0.681 = 6.13 \, \text{cm}
\]
Bài 64:
\[
BC^2 = AB^2 + AC^2
\]
\[
BC^2 = 3^2 + 4^2 = 9 + 16 = 25
\]
\[
BC = \sqrt{25} = 5 \, \text{cm}
\]
\[
\tan(B) = \frac{AC}{AB} = \frac{4}{3} \implies B = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ
\]
\[
C = 90^\circ - B \approx 90^\circ - 53.13^\circ \approx 36.87^\circ
\]
Bài 65:
\[
BC^2 = AB^2 + AC^2
\]
\[
10^2 = 6^2 + AC^2
\]
\[
100 = 36 + AC^2
\]
\[
AC^2 = 100 - 36 = 64 \implies AC = \sqrt{64} = 8 \, \text{cm}
\]
\[
\tan(B) = \frac{AC}{AB} = \frac{8}{6} = \frac{4}{3} \implies B = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ
\]
\[
C = 90^\circ - B \approx 90^\circ - 53.13^\circ \approx 36.87^\circ
\]
