cho tam giác ABC có BC=a CA=b AB=c, thỏa mãn (a+b+c)(a+b-c)=3ab. Tìm độ lớn góc C?
cho tam giác ABC có BC=a CA=b AB=c, thỏa mãn (a+b+c)(a+b-c)=3ab. Tìm độ lớn góc C?
\(\left(a+b+c\right)\left(a+b-c\right)=3ab\)
\(\Leftrightarrow\left(a+b\right)^2-c^2=3ab\)
\(\Leftrightarrow a^2+b^2+2ab-c^2=3ab\)
\(\Leftrightarrow a^2+b^2-c^2=ab\)
\(\Leftrightarrow\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{1}{2}\)
\(\Rightarrow cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{1}{2}\)
\(\Rightarrow C=60^0\)
cho tanx=2 tính Q=\(\dfrac{\sin^3x}{2\sin x+\cos^3x}\)
tan x=2
=>\(\dfrac{sinx}{cosx}=2\)
=>sin x và cosx cùng dấu và \(sinx=2\cdot cosx\)
\(1+tan^2x=\dfrac{1}{cos^2x}\)
=>\(\dfrac{1}{cos^2x}=1+4=5\)
=>\(cos^2x=\dfrac{1}{5}\)
=>\(\left[{}\begin{matrix}cosx=\dfrac{1}{\sqrt{5}}\\cosx=-\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)
TH1: \(cosx=\dfrac{1}{\sqrt{5}}\)
=>\(sinx=\sqrt{1-cos^2x}=\dfrac{2}{\sqrt{5}}\)
TH2: cosx=-1/căn 5
=>\(sinx=-\sqrt{1-cos^2x}=-\dfrac{2}{\sqrt{5}}\)
\(Q=\dfrac{sin^3x}{2sinx+cos^3x}\)
\(=\dfrac{\left(2\cdot cosx\right)^3}{2\cdot2cosx+cos^3x}\)
\(=\dfrac{8\cdot cos^3x}{4cosx+cos^3x}=\dfrac{8cos^2x}{4+cos^2x}\)
\(=\dfrac{8\cdot\dfrac{1}{5}}{4+\dfrac{1}{5}}=\dfrac{8}{5}:\dfrac{21}{5}=\dfrac{8}{21}\)
Cho cotx=2 . Tính giá trị của biểu thức B= sin^ 2 x-2 sin x.cos x-1 / 5cos^2 x + sin^2 x - 3
cotx=2
=>cosx=2*sin x
\(1+cot^2x=\dfrac{1}{sin^2x}\)
=>\(\dfrac{1}{sin^2x}=1+4=5\)
=>\(sin^2x=\dfrac{1}{5}\)
\(B=\dfrac{sin^2x-2\cdot sinx\cdot2\cdot sinx-1}{5\cdot4sin^2x+sin^2x-3}=\dfrac{-3sin^2x-1}{21sin^2x-3}\)
\(=\dfrac{-\dfrac{3}{5}-1}{\dfrac{21}{5}-3}=-\dfrac{8}{5}:\dfrac{6}{5}=-\dfrac{4}{3}\)
\(cotx=2\Rightarrow tanx=\dfrac{1}{2}\)
\(B=\dfrac{sin^2x-2sinx.cosx-1}{5cos^2x+sin^2x-3}\)
\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-\dfrac{1}{cos^2x}}{5+tan^2x-\dfrac{3}{cos^2x}}\)
\(\Leftrightarrow B=\dfrac{tan^2x-2tanx-1-tan^2x}{5+tan^2x-3-3tan^2x}\)
\(\Leftrightarrow B=\dfrac{-2tanx-1}{2-2tan^2x}\)
\(\Leftrightarrow B=\dfrac{-2.\dfrac{1}{2}-1}{2-2.\dfrac{1}{4}}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)
Chứng minh ràng trong tam giác ABC sin (A + 2B)/2 = cos((C - B)/2)
Ta có: `\hat{A}+\hat{B}+\hat{C}=180^o`
`=>\hat{C}-\hat{B}=180^o-\hat{A}-2\hat{B}`
`=>[\hat{C}-\hat{B}]/2=90^o - [\hat{A}+2\hat{B}]/2`
`=>sin` `[\hat{A}+2\hat{B}]/2 = cos` `[\hat{C}-\hat{B}]/2`
`=>đpcm`.
Cho tam giác ABC có cạnh AB =4,2 BC=7,2 CA=6,5 Đường trung tuyến AM cắt phân giác BD tại I .Tính gần đung IA,IB
\(AM=\sqrt{\dfrac{4.2^2+6.5^2}{2}-\dfrac{7.2^2}{4}}\simeq4,12\left(cm\right)\)
BM=BC/2=3,6cm
Xét ΔBAM có BI là phân giác
nên IA/AB=IM/BM
=>IA/42=IM/36
=>IA/7=IM/6=(IA+IM)/(7+6)=AM/13=4,12/13
=>\(IA\simeq2,22\left(cm\right);IM\simeq1,9\left(cm\right)\)
\(\dfrac{sin^2x+3sinx.cosx+1}{2cos^2x-3}\)
\(=\left(\dfrac{1-cos2x}{2}+\dfrac{3\cdot sin2x}{2}+1\right):\left(2\cdot\dfrac{1+cos2x}{2}-3\right)\)
\(=\dfrac{1-cos2x+3sin2x+2}{2}:\dfrac{2+2cos2x-6}{2}\)
\(=\dfrac{3sin2x-cos2x+3}{2cos2x-4}\)
cho a4=b4 + c4. CMR: 2sin2A=tanB.tanC
Cho tam giác ABC có:(Aa+Bb+Cc)/(a+b+c)=(A+B+C)/3 CM tam giác ABC đều
a.
\(BC=\sqrt{AB^2+AC^2-2AB.AC.cosA}=10,28\)
Áp dụng định lý hàm sin:
\(\dfrac{BC}{sinA}=2R\Rightarrow R=\dfrac{BC}{2sinA}=7,27\)
\(S=\dfrac{1}{2}Ab.AC.sinA=9,28\)
\(p=\dfrac{AB+AC+BC}{2}=10,64\)
\(\Rightarrow r=\dfrac{S}{p}=0,87\)
b.
Áp dụng định lý hàm cos:
\(BC^2=AB^2+AC^2-2AB.AC.cosA\)
\(\Rightarrow7^2=5^2+AC^2-2.5.AC.cos60^0\)
\(\Rightarrow AC^2-5AC-24=0\)
\(\Rightarrow AC=8\)
Áp dụng định lý hàm sin:
\(R=\dfrac{BC}{2sinA}=\dfrac{7\sqrt{3}}{3}\)
\(p=\dfrac{AB+AC+BC}{2}=10\)
\(S=\dfrac{1}{2}AB.AC.sinA=10\sqrt{3}\)
\(\Rightarrow r=\dfrac{S}{p}=\sqrt{3}\)
\(h_a=\dfrac{2S}{BC}=\dfrac{20\sqrt{3}}{7}\)
c.
\(p=\dfrac{a+b+c}{2}=\dfrac{21}{2}\)
\(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=21\sqrt{130}\)
\(h_a=\dfrac{2S}{a}=6\sqrt{130}\)
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