Giải tam giác ABC vuông tại A, biết rằng:

a) \[b = 10cm,\,\widehat {\,C} = 30^\circ ;\]         b) \[c = 10cm,\,\,\widehat C = 45^\circ ;\]

c) \({\rm{a}} = 20\,\;{\rm{cm}},\,\,\widehat {\rm{B}} = 35^\circ \);                                                               d) \({\rm{c}} = 21{\rm{cm}},{\rm{b}} = 18\;{\rm{cm}}\).

a) (h.28)

\(\widehat {\rm{B}} = 90^\circ - 30^\circ = 60^\circ \);

\({\rm{AB}} = {\rm{AC}} \cdot \tan {\rm{C}} = 10 \cdot \tan 30^\circ \approx 5,774(\;{\rm{cm}})\);

$\text{BC}=\frac{\text{AC}}{\cos \text{C}}=\frac{10}{\cos {30}^{°}}\approx 11,547( \text{cm}).$

b) (h.29)

\(\widehat {\rm{B}} = 90^\circ - 45^\circ = 45^\circ \);

\({\rm{AC}} = {\rm{AB}} \cdot \cot {\rm{C}} = 10 \cdot \cot 45^\circ = 10\,\,(\;{\rm{cm}})\).

\({\rm{BC}} = \frac{{{\rm{AB}}}}{{\sin {\rm{C}}}} = \frac{{10}}{{\sin 45^\circ }} \approx 14,142(\;{\rm{cm}}).\)

c) (h.30)

\(\widehat {\rm{C}} = 90^\circ - 35^\circ = 55^\circ \);

\({\rm{AB}} = {\rm{BC}} \cdot \cos {\rm{B}} = 20 \cdot \cos 35^\circ \approx 16,383\,\,({\rm{cm}})\);

\({\rm{AC}} = {\rm{BC}} \cdot \sin {\rm{B}} = 20 \cdot \sin 35^\circ \approx 11,472\,\,(\;{\rm{cm}})\).

d) (h.31)

\(\tan B = \frac{{AC}}{{AB}} = \frac{{18}}{{21}} \approx 0,8571;\)

\( \Rightarrow \widehat {\rm{B}} \approx 41^\circ \) và \(\widehat {\rm{C}} = 49^\circ \);

\({\rm{BC}} = \frac{{{\rm{AC}}}}{{\sin {\rm{B}}}} = \frac{{18}}{{\sin 41^\circ }} \approx 27,437\,\,(\;{\rm{cm}}).\)