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Ta có : \(\widehat{QTS}=180^0-\widehat{QTP}=180^0-150^0=30^0\)

Trong tam giác vuông QST, ta có:

\(QS=QT.sinQTS=8.sin30^0=4\left(cm\right)\)

\(TS=QT.cosQTS=8.cos30^0\sim6,928\left(cm\right)\)

Trong tam giác vuông QSP, ta có:

\(SP=QS.cotQPS=4.cot18^0=12,311\left(cm\right)\)

\(PT=SP-TS\sim12,311-6,928\sim5,383\left(cm\right)\)

b) Ta có:

\(S_{QPR}=\frac{1}{2}.QS.PR=\frac{1}{2}.QS.\left(PT+TR\right)\sim\frac{1}{2}.4.\left(5,383+5\right)\sim20,766\left(cm^2\right)\)

Ta có: \(a = BC = 20;\;b = AC = 15;\;c = AB = 12.\)

a) Áp dụng định lí cosin trong tam giác ABC, ta có:

 \(\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}};\;\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\)

\( \Rightarrow \cos A = \frac{{{{15}^2} + {{12}^2} - {{20}^2}}}{{2.15.12}};\;\cos B = \frac{{{{20}^2} + {{12}^2} - {{15}^2}}}{{2.20.12}}\)

\( \Rightarrow \cos A =  - \frac{{31}}{{360}};\;\cos B = \frac{{319}}{{480}}\)

\( \Rightarrow \widehat A = 94,{9^o};\;\widehat B = 48,{3^o}\)

\( \Rightarrow \widehat C = {180^o} - \left( {94,{9^o} + 48,{3^o}} \right) = 36,{8^o}\)

b)

Diện tích tam giác ABC là: \(S = \frac{1}{2}.bc.\sin A = \frac{1}{2}.15.12.\sin 94,{9^o} \approx 89,7.\)

Ko thấy hình ảnh luôn :>>>>