Đáp án:
Áp dụng hệ thức lượng, định lý Cos trong tam giác ta có:
$\begin{array}{l}
1)a)\cos \widehat A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\\
= \dfrac{{{{15}^2} + {{13}^2} - {{12}^2}}}{{2.15.13}}\\
= \dfrac{{25}}{{39}}\\
\Rightarrow \widehat A = {50^0}\\
\cos \widehat B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2.a.c}} = \dfrac{{11}}{{39}}\\
\Rightarrow \widehat B = {73^0}\\
\Rightarrow \widehat C = {180^0} - \widehat A - \widehat B = {57^0}\\
b)A{M^2} = \dfrac{{{b^2} + {c^2}}}{2} - \dfrac{{{a^2}}}{4} = 161\\
\Rightarrow AM = \sqrt {161} \\
B{N^2} = \dfrac{{{a^2} + {c^2}}}{2} - \dfrac{{{b^2}}}{4} = \dfrac{{401}}{4}\\
\Rightarrow BN = \dfrac{{\sqrt {401} }}{2}\\
C{P^2} = \dfrac{{{a^2} + {b^2}}}{2} - \dfrac{{{c^2}}}{4} = \dfrac{{569}}{4}\\
\Rightarrow CP = \dfrac{{\sqrt {569} }}{2}\\
c)p = \dfrac{{a + b + c}}{2} = 20\\
\Rightarrow S = \sqrt {p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)} \\
= \sqrt {20.\left( {20 - 12} \right).\left( {20 - 15} \right).\left( {20 - 13} \right)} \\
= 20\sqrt {14} \\
S = \dfrac{{abc}}{{4R}} = p.r\\
\Rightarrow \left\{ \begin{array}{l}
R = \dfrac{{abc}}{{4S}} = \dfrac{{117\sqrt {14} }}{{56}}\\
r = \dfrac{S}{p} = \sqrt {14}
\end{array} \right.\\
d)\left\{ \begin{array}{l}
{h_a} = \dfrac{{2S}}{a} = \dfrac{{40\sqrt {14} }}{{12}} = \dfrac{{10\sqrt {14} }}{3}\\
{h_B} = \dfrac{{40\sqrt {14} }}{{15}} = \dfrac{{8\sqrt {14} }}{3}\\
{h_C} = \dfrac{{40\sqrt {14} }}{{13}}
\end{array} \right.\\
2)a){S_{ABC}} = \dfrac{1}{2}.AB.AC.\sin \widehat A\\
= \dfrac{1}{2}.6.8.\sin {120^0}\\
= 12\sqrt 3 \\
b)B{C^2} = A{B^2} + A{C^2} - 2.AB.AC.\cos \widehat A\\
= {6^2} + {8^2} - 2.6.8.\cos {120^0}\\
= 148\\
\Rightarrow BC = 2\sqrt {37} \\
Do:S = \dfrac{{AB.AC.BC}}{{4R}}\\
\Rightarrow R = \dfrac{{4S}}{{AB.AC.BC}} = \dfrac{{4.12\sqrt 3 }}{{6.8.2\sqrt {37} }} = \dfrac{{\sqrt {111} }}{{74}}\\
3)a)\cos \widehat C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2.a.b}}\\
= \dfrac{{{8^2} + {{10}^2} - {{13}^2}}}{{2.8.10}} = - \dfrac{1}{{32}}\\
\Rightarrow \widehat C = {91^0}47' > {90^0}
\end{array}$
=> Tam giác ABC có góc tù tại C
$\begin{array}{l}
b)a = 8;b = 10;c = 13\\
\Rightarrow p = \dfrac{{a + b + c}}{2} = \dfrac{{31}}{2}\\
\Rightarrow S = \sqrt {p\left( {p - a} \right).\left( {p - b} \right).\left( {p - c} \right)} \\
= 519\\
\Rightarrow R = \dfrac{{4S}}{{a.b.c}} = 1,996\\
c)S = 519\\
4)\widehat C = {180^0} - {60^0} - {45^0} = {75^0}\\
\dfrac{a}{{\sin \widehat A}} = \dfrac{b}{{\sin \widehat B}} = \dfrac{c}{{\sin \widehat C}} = 2R\\
\Rightarrow \left\{ \begin{array}{l}
a = \sqrt 6 \\
c = 1 + \sqrt 3 \\
R = \sqrt 2
\end{array} \right.\\
\Rightarrow S = \dfrac{1}{2}.a.b.\sin \widehat C = \dfrac{{3 + \sqrt 3 }}{2}
\end{array}$
