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\(cosA=\dfrac{AB^2+AC^2-BC^2}{2AB.AC}=-\dfrac{1}{32}\)
\(\Rightarrow A\approx92^0\)
\(p=\dfrac{AB+AC+BC}{2}=\dfrac{31}{2}\)
\(S_{ABC}=\sqrt{p\left(p-AB\right)\left(p-AC\right)\left(p-BC\right)}\simeq40\)
\(r=\dfrac{S}{p}=\dfrac{80}{31}\)
Tham khảo:
a) Áp dụng hệ quả của định lí cosin, ta có:
\(\begin{array}{l}\cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}};\cos B = \frac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\\ \Rightarrow \left\{ \begin{array}{l}\cos A = \frac{{{{10}^2} + {{13}^2} - {8^2}}}{{2.10.13}} = \frac{{41}}{{52}} > 0;\\\cos B = \frac{{{8^2} + {{13}^2} - {{10}^2}}}{{2.8.13}} = \frac{{133}}{{208}} > 0\\\cos C = \frac{{{8^2} + {{10}^2} - {{13}^2}}}{{2.8.10}} = - \frac{1}{{32}} < 0\end{array} \right.\end{array}\)
\( \Rightarrow \widehat C \approx 91,{79^ \circ } > {90^ \circ }\), tam giác ABC có góc C tù.
b)
+) Áp dụng định lí cosin trong tam giác ACM, ta có:
\(\begin{array}{l}A{M^2} = A{C^2} + C{M^2} - 2.AC.CM.\cos C\\ \Leftrightarrow A{M^2} = {8^2} + {5^2} - 2.8.5.\left( { - \frac{1}{{32}}} \right) = 91,5\\ \Rightarrow AM \approx 9,57\end{array}\)
+) Ta có: \(p = \frac{{8 + 10 + 13}}{2} = 15,5\).
Áp dụng công thức heron, ta có: \(S = \sqrt {p(p - a)(p - b)(p - c)} = \sqrt {15,5.(15,5 - 8).(15,5 - 10).(15,5 - 13)} \approx 40\)
+) Áp dụng định lí sin, ta có:
\(\frac{c}{{\sin C}} = 2R \Rightarrow R = \frac{c}{{2\sin C}} = \frac{{13}}{{2.\sin 91,{{79}^ \circ }}} \approx 6,5\)
c)
Ta có: \(\widehat {BCD} = {180^ \circ } - 91,{79^ \circ } = 88,{21^ \circ }\); \(CD = AC = 8\)
Áp dụng định lí cosin trong tam giác BCD, ta có:
\(\begin{array}{l}B{D^2} = C{D^2} + C{B^2} - 2.CD.CB.\cos \widehat {BCD}\\ \Leftrightarrow B{D^2} = {8^2} + {10^2} - 2.8.10.\cos 88,{21^ \circ } \approx 159\\ \Rightarrow BD \approx 12,6\end{array}\)
\(BC=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos A=148\left(cm\right)\)
Đặt AB = c ; AC = b ; BC = a .
Ta có : \(b+c=13\) ; \(r=\dfrac{S}{p}=\sqrt{3}\) ( p \(=\dfrac{a+b+c}{2}\) )
Có : \(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\) nên : \(r=\sqrt{\dfrac{\left(p-a\right)\left(p-b\right)\left(p-c\right)}{p}}=\sqrt{3}\)
\(\Rightarrow\left(p-a\right)\left(p-b\right)\left(p-c\right)=3p\)
\(\Leftrightarrow\left(\dfrac{-a+b+c}{2}\right)\left(\dfrac{-b+a+c}{2}\right)\left(\dfrac{-c+a+b}{2}\right)=\dfrac{3\left(a+b+c\right)}{2}\)
\(\Leftrightarrow\left(-a+b+c\right)\left(-b+a+c\right)\left(-c+a+b\right)=12\left(a+b+c\right)\)
\(\Leftrightarrow\left(-a+13\right)\left(-b+a+c\right)\left(-c+a+b\right)=12\left(13+a\right)\)
\(\Leftrightarrow\left(-a+13\right)\left[a^2-\left(b-c\right)^2\right]=12\left(13+a\right)\) (2)
Có : \(\dfrac{b^2+c^2-a^2}{2bc}=cosA=cos60^o=\dfrac{1}{2}\) \(\Rightarrow b^2+c^2-a^2=bc\) \(\Leftrightarrow a^2=b^2+c^2-bc\) (1)
Mặt khác : \(b+c=13\Leftrightarrow b^2+c^2-bc+3bc=169\Leftrightarrow a^2=169-3bc\)
Từ (1) ; (2) suy ra : \(\left(-a+13\right)bc=12\left(13+a\right)\)
\(\Leftrightarrow\left(-a+13\right)\left(169-a^2\right)=36\left(13+a\right)\)
\(\Leftrightarrow\left(13-a\right)^2\left(13+a\right)=36\left(13+a\right)\)
\(\Leftrightarrow\left(13-a\right)^2=36\) \(\Leftrightarrow\left[{}\begin{matrix}13-a=6\\13-a=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=7\\a=19>13=b+c\left(L\right)\end{matrix}\right.\)
Vậy ...
\(\cos A=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{8^2+10^2-13^2}{2\cdot8\cdot10}=-\dfrac{1}{32}< 0\)
nên \(\widehat{A}>90^0\)
=>ΔABC tù