§3. Các hệ thức lượng giác trong tam giác và giải tam giác

D.3.8 cm

d

D

\(4m_a^2=b\left(b+4c.cosA\right)=b^2+4bc.cosA\)

\(\Leftrightarrow4\left(\dfrac{2b^2+2c^2-a^2}{4}\right)=b^2+4bc.\dfrac{b^2+c^2-a^2}{2bc}\)

\(\Leftrightarrow2b^2+2c^2-a^2=b^2+2\left(b^2+c^2-a^2\right)\)

\(\Leftrightarrow a^2=b^2\)

\(\Leftrightarrow a=b\)

\(\Rightarrow\Delta ABC\) cân tại C

Tham khảo:

chọn C = 60 độ á

a) Xét \(\Delta ABC:\)

\(\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{7^2+5^2-\left(4\sqrt{2}\right)^2}{2.7.5}=\dfrac{3}{5}.\\ \cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{\left(4\sqrt{2}\right)^2+5^2-7^2}{2.4\sqrt{2}.5}=\dfrac{\sqrt{2}}{20}.\\ \cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{\left(4\sqrt{2}\right)^2+7^2-5^2}{2.4\sqrt{2}.7}=\dfrac{\sqrt{2}}{2}.\)

b) Xét \(\Delta MBN:\)

\(MN^2=MB^2+BN^2-2.MB.BN.\cos B.\\ \Rightarrow MN^2=\left(\dfrac{1}{3}AB\right)^2+\left(\dfrac{1}{2}BC\right)^2-2.\dfrac{1}{3}AB.\dfrac{1}{2}BC.\dfrac{\sqrt{2}}{20}.\\ =\dfrac{1}{9}.5^2+\dfrac{1}{4}.\left(4\sqrt{2}\right)^2-\dfrac{2}{3}.5.\dfrac{1}{2}.4\sqrt{2}.\dfrac{\sqrt{2}}{20}.\\ =\dfrac{91}{9}.\)

c) Ta có:

 \(\cos A=\dfrac{3}{5}.\\ \Rightarrow\widehat{A}\approx53,13^o.\\ \Rightarrow\sin A=\dfrac{4}{5}.\)

 \(S_{\Delta BMN}=\dfrac{1}{2}bc\sin A.\\ =\dfrac{1}{2}.7.5.\dfrac{4}{5}=14\text{(đvdt).}\) 

a: \(\cos A=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{10+36-34}{2\cdot\sqrt{10}\cdot6}=\dfrac{\sqrt{10}}{10}\)

b: AM=2/3AC=4

Xét ΔABM có 

\(\cos A=\dfrac{AB^2+AM^2-BM^2}{2\cdot AB\cdot AM}=\dfrac{10+16-BM^2}{2\cdot\sqrt{10}\cdot4}=\dfrac{\sqrt{10}}{10}\)

\(\Leftrightarrow26-BM^2=8\)

hay \(BM=3\sqrt{2}\)

\(\cos A=\sqrt{1-\dfrac{16}{25}}=\dfrac{3}{5}\)

\(\cos A=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(\Leftrightarrow74-BC^2=\dfrac{4}{5}\cdot2\cdot5\cdot7=56\)

hay \(BC=3\sqrt{2}\)