Bài 3. CÁC HỆ THỨC LƯỢNG TRONG TAM GIÁC VÀ GIẢI TAM GIÁC

Gọi G là giao điểm BM và CN. Đặt AB=c, AC=b

Ta có: \(BM^2=\dfrac{2\left(a^2+c^2\right)-b^2}{4}\) ; \(\Rightarrow BG^2=\left(\dfrac{2}{3}BM\right)^2=\dfrac{2\left(a^2+c^2\right)-b^2}{9}\)

\(CN^2=\dfrac{2\left(a^2+b^2\right)-c^2}{4}\Rightarrow CG^2=\dfrac{2\left(a^2+b^2\right)-c^2}{9}\)

Mặt khác \(BG^2+CG^2=BC^2\)

\(\Rightarrow\dfrac{2\left(a^2+c^2\right)-b^2}{9}+\dfrac{2\left(a^2+b^2\right)-c^2}{9}=a^2\)

\(\Rightarrow b^2+c^2=5a^2\)

Áp dụng định lý hàm cos:

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{5a^2-a^2}{2bc}=\dfrac{2a^2}{bc}\Rightarrow bc=\dfrac{2a^2}{cos\alpha}\)

\(S_{ABC}=\dfrac{1}{2}bcsinA=\dfrac{1}{2}.\dfrac{2a^2}{cos\alpha}.sin\alpha=a^2.tan\alpha\)

\(AC=\sqrt{AB^2+BC^2-2AB.BC.cosB}=\sqrt{9^2+12^2-2.9.12.cos60^0}=3\sqrt{13}\)

\(C=180^0-\left(A+B\right)=75^0\)

Áp dụng định lý hàm sin:

\(\dfrac{b}{sinB}=\dfrac{c}{sinC}\Rightarrow c=\dfrac{b.sinC}{sinB}=\dfrac{8.sin75^0}{sin45^0}=4+4\sqrt{3}\)

\(BM=\dfrac{1}{2}BC=3\)

\(AM=\sqrt{AB^2+BM^2-2AB.BM.cos60^0}=\sqrt{19}\)

\(BN=\dfrac{\sqrt{2\left(AB^2+BM^2\right)-AM^2}}{2}=\dfrac{7}{2}\)

\(\widehat{A}=180^o-\widehat{D}=180^o-60^o=120^o\)

\(S_{ABD}=\dfrac{1}{2}.AB.AD.sin120^o=10\sqrt{3}\left(đvdt\right)\)

\(\Rightarrow S_{ABCD}=2S_{ABD}=20\sqrt{3}\left(đvdt\right)\)

Theo định lí hàm số cosin:

\(a^2=b^2+c^2-2bc.cosA=6^2+4^2-2.6.4.cos120^o=76\Rightarrow a=2\sqrt{19}\)

\(cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{76+16-36}{2.2\sqrt{19}.4}=\dfrac{7\sqrt{19}}{38}\)

\(cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{76+36-16}{2.2\sqrt{19}.6}=\dfrac{4\sqrt{19}}{19}\)

a² = b² + c² - 2bc.cos120⁰

⇔ a² = 76

⇔ a = \(2\sqrt{19}\)

Theo định lí sin: \(\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}\)

⇔ \(\dfrac{2\sqrt{19}}{sin120}=\dfrac{6}{sinB}=\dfrac{4}{sinC}\)

⇔ \(\left\{{}\begin{matrix}sinC=\dfrac{\sqrt{57}}{19}\\sinB=\dfrac{3\sqrt{57}}{38}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{B}=36^035'\\\widehat{C}=23^025'\end{matrix}\right.\) 

Lời giải:

Có vẻ đề thiếu dữ kiện độ dài $AC$.

Bạn chỉ cần nhớ công thức:

\(\cos \widehat{BAC}=\cos (\overrightarrow{AB}, \overrightarrow{AC})=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}|.|\overrightarrow{AC}|}=\cos 120=\frac{-1}{2}\)

\(\Rightarrow \overrightarrow{AB}.\overrightarrow{AC}=\frac{-1}{2}.|\overrightarrow{AB}|.|\overrightarrow{AC}|=\frac{-1}{2}.AB.AC=\frac{-1}{2}.10.AC\)

Đến đây bạn thay giá trị của $AC$ vào nữa để tính.

\(cosA+cosB+cosC=2cos\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{A-B}{2}\right)+1-2sin^2\dfrac{C}{2}\)

\(=-2sin^2\dfrac{C}{2}+2sin\dfrac{C}{2}cos\left(\dfrac{A-B}{2}\right)+1\)

\(=-2\left[sin\dfrac{C}{2}-\dfrac{1}{2}cos\dfrac{A-B}{2}\right]^2-\dfrac{1}{2}sin^2\dfrac{A-B}{2}+\dfrac{3}{2}\le\dfrac{3}{2}\)

\(\left\{{}\begin{matrix}\dfrac{a+b}{6}=\dfrac{b+c}{5}\\\dfrac{a+b}{6}=\dfrac{c+a}{7}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=\dfrac{a}{2}\\c=\dfrac{3a}{4}\end{matrix}\right.\)

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{\dfrac{a^2}{4}+\dfrac{9a^2}{16}-a^2}{2.\dfrac{a}{2}.\dfrac{3a}{4}}=-\dfrac{1}{4}\)

\(cosB=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{a^2+\dfrac{9a^2}{16}-\dfrac{a^2}{4}}{2a.\dfrac{3a}{4}}=\dfrac{7}{8}\)

\(cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{11}{16}\)

\(P=-\dfrac{1}{4}+\dfrac{14}{8}+\dfrac{44}{16}=\dfrac{17}{4}\)