a) (H.4.31)

Ta có:

\(\widehat B = 90^\circ - \widehat C = 90^\circ - 40^\circ = 50^\circ ,\)

\(AC = BC.\cos C = 20.\cos 40^\circ \approx 15,3,\)

\(AB = BC.\sin C = 20.\sin 40^\circ \approx 12,8.\)

b) (H.4.32)

Ta có

\(\widehat C = 90^\circ - \widehat B = 90^\circ - 55^\circ = 35^\circ ,\)

\(AB = AC.\cot B = 82.\cot 55^\circ \approx 57,4,\)

\(B{C^2} = A{B^2} + A{C^2} = {57,4^2} + {82^2} = 10018,76,\) suy ra \(BC = \sqrt {10\,\,018,76} \approx 100,1.\)

c) (H.4.33)

Ta có

\(A{B^2} = B{C^2} - A{C^2} = {32^2} - {20^2} = 624,\) suy ra \(AB = \sqrt {624} \approx 25.\)

\(\sin B = \frac = \frac = 0,625,\) suy ra \(\widehat B \approx 39^\circ .\)

Từ đó suy ra \(\widehat C = 90^\circ - \widehat B = 90^\circ - 39^\circ = 51^\circ .\)