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1: A(1;1); B(3;3); C(0;-6)
\(AB=\sqrt{\left(3-1\right)^2+\left(3-1\right)^2}=\sqrt{2^2+2^2}=2\sqrt2\)
\(AC=\sqrt{\left(0-1\right)^2+\left(-6-1\right)^2}=\sqrt{\left(-1\right)^2+\left(-7\right)^2}=\sqrt{50}=5\sqrt2\)
\(BC=\sqrt{\left(0-3\right)^2+\left(-6-3\right)^2}=\sqrt{\left(-3\right)^2+\left(-9\right)^2}=\sqrt{90}=3\sqrt{10}\)
Xét ΔABC có \(cosA=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
\(=\frac{8+50-90}{2\cdot2\sqrt2\cdot5\sqrt2}=\frac{8-40}{4\cdot2\cdot5}=\frac{-32}{8\cdot5}=\frac{-4}{5}\)
2: D(x;y); A(1;1); B(3;3)
\(\overrightarrow{DA}=\left(1-x;1-y\right);\overrightarrow{DB}=\left(3-x;3-y\right)\)
ΔDAB vuông cân tại D
=>DA=DB và \(\overrightarrow{DA}\cdot\overrightarrow{DB}=0\)
DA=DB
=>\(\left(1-x\right)^2+\left(1-y\right)^2=\left(3-x\right)^2+\left(3-y\right)^2\)
=>\(x^2-2x+1+y^2-2y+1=x^2-6x+9+y^2-6y+9\)
=>-2x-2y+2=-6x-6y+18
=>4x+4y=16
=>x+y=4
=>y=4-x
\(\overrightarrow{DA}\cdot\overrightarrow{DB}=0\)
=>(1-x)(3-x)+(1-y)(3-y)=0
=>(x-1)(x-3)+(y-1)(y-3)=0
=>(x-1)(x-3)+(4-x-1)(4-x-3)=0
=>(x-1)(x-3)+(3-x)(1-x)=0
=>2(x-1)(x-3)=0
=>(x-1)(x-3)=0
=>x=1 hoặc x=3
TH1: x=1
=>y=4-x=4-1=3
=>D(1;3)
TH2: x=3
=>y=4-x=4-3=1
=>D(3;1)
a: A(-2;6); B(1;2); C(9;8)
\(AB=\sqrt{\left(1+2\right)^2+\left(2-6\right)^2}=\sqrt{3^2+\left(-4\right)^2}=5\)
\(AC=\sqrt{\left(9+2\right)^2+\left(8-6\right)^2}=\sqrt{11^2+2^2}=\sqrt{125}=5\sqrt5\)
\(BC=\sqrt{\left(9-1\right)^2+\left(8-2\right)^2}=\sqrt{8^2+6^2}=10\)
Vì \(BA^2+BC^2=AC^2\)
nên ΔABC vuông tại B
b: Tọa độ I là:
\(\begin{cases}x_{I}=\frac12\cdot\left(x_{A}+x_{C}\right)=\frac12\cdot\left(-2+9\right)=\frac72\\ y_{I}=\frac12\cdot\left(y_{A}+y_{C}\right)=\frac12\cdot\left(6+8\right)=\frac12\cdot14=7\end{cases}\)
=>I(7/2;7)
I(7/2;7); A(-2;6); B(1;2); H(x;y)
H là trực tâm cua ΔIAB
=>IH⊥AB và AH⊥BI
\(\overrightarrow{AH}=\left(x+2;y-6\right);\overrightarrow{BI}=\left(\frac72-1;7-2\right)=\left(\frac52;5\right)\)
AH⊥BI
=>\(\overrightarrow{AH}\cdot\overrightarrow{BI}=0\)
=>\(\frac52\left(x+2\right)+5\left(y-6\right)=0\)
=>\(\frac12\left(x+2\right)+\left(y-6\right)=0\)
=>x+2+2(y-6)=0
=>x+2+2y-12=0
=>x+2y-10=0
=>x=-2y+10
IH⊥AB
=>\(\overrightarrow{IH}\cdot\overrightarrow{AB}=0\)
mà \(\overrightarrow{IH}=\left(x-\frac72;y-7\right);\overrightarrow{AB}=\left(1+2;2-6\right)=\left(3;-4\right)\)
nên \(3\left(x-\frac72\right)+\left(-4\right)\left(y-7\right)=0\)
=>3x-10,5-4y+28=0
=>3x-4y+17,5=0
=>3(-2y+10)-4y+17,5=0
=>-6y+30-4y+17,5=0
=>-10y+47,5=0
=>-10y=-47,5
=>y=4,75
x=-2y+10=-2*4,75+10=-9,5+10=0,5
=>H(0,5;4,75)
Bài 3: Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>a+b+c=180
Ta có: \(\hat{C}-3\cdot\hat{B}-2\cdot\hat{A}=-3^0\)
=>c-3b-2a=-3
=>2a+3b-c=3
mà a+b+c=180
nên 2a+3b-c+a+b+c=3+180
=>3a+4b=183
=>6a+8b=366
\(5\cdot\hat{B}-2\cdot\hat{A}=16^0\)
=>5b-2a=16
=>15b-6a=48
=>15b-6a+6a+8b=366+48
=>23b=414
=>\(b=\frac{414}{23}=18^0\)
=>\(\hat{B}=18^0\)
3a+4b=183
=>3a=183-4b=183-72=111
=>\(a=\frac{111}{3}=37^0\)
=>\(\hat{A}=37^0\)
\(\hat{C}=180^0-18^0-37^0=180^0-55^0=125^0\)
Bài 2:
Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>a+b+c=180
\(\hat{A}+\hat{B}-2\cdot\hat{C}=27^0\)
=>a+b-2c=27
=>(a+b+c)-(a+b-2c)=180-27
=>3c=153
=>\(c=\frac{153}{3}=51\)
=>\(\hat{C}=51^0\)
\(\hat{A}+3\cdot\hat{C}=273^0\)
=>\(\hat{A}=273^0-3\cdot51^0=273^0-153^0=120^0\)
\(\hat{B}=180^0-51^0-120^0=60^0-51^0=9^0\)
bài 1:
Đặt \(\hat{A}=a;\hat{B}=b;\hat{C}=c\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>a+b+c=180
\(\hat{A}-\hat{B}+\hat{C}=90^0\)
=>a-b+c=90
=>a+b+c-(a-b+c)=180-90
=>2b=90
=>b=45
=>\(\hat{B}=45^0\)
=>\(\hat{A}+\hat{C}=180^0-45^0=135^0\)
mà \(\hat{A}-\hat{C}=-5^0\)
nên \(\hat{A}=\frac{135^0-5^0}{2}=\frac{130^0}{2}=65^0\)
=>\(\hat{C}=135^0-65^0=70^0\)
