Cho tam giác ABC có A(3; -3), B(-3; 5), C(3; 5). Tâm đường tròn ngoại tiếp của tam giác ABC có tọa độ là
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
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 \(\hat{A}=\frac{135^0-5^0}{2}=\frac{130^0}{2}=65^0\)
=>\(\hat{C}=135^0-65^0=70^0\)
bài 2:
ta có: AB<AC<BC(Vì 3cm<4cm<5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
bài 2:
ta có: AB <AC <BC (Vì 3cm <4cm <5cm)
=> góc C>góc A> góc B (Các cạnh và góc đồi diện trong tam giác)
Bài 3:
*Xét tam giác ABC, có:
góc A+góc B+góc c= 180 độ( tổng 3 góc 1 tam giác)
hay góc A+60 độ +40 độ=180độ
=> góc A= 180 độ-60 độ-40 độ.
=> góc A=80 độ
Ta có: góc A>góc B>góc C(vì 80 độ>60 độ>40 độ)
=> BC>AC>AB( Các cạnh và góc đối diện trong tam giác)
HT mik làm giống bạn Dương Mạnh Quyết
Chọn B.
Ta có:

Mặt khác ![]()
Suy ra diện tích tam giác ABC là 1/2.AB.BC = 6.
Gọi I(a;b) là tâm đường tròn ngoại tiếp tam giác ABC.
Ta có: AI = BI = CI ⇔ AI2 = BI2 = CI2
A I 2 = B I 2 B I 2 = C I 2 ⇔ a − 3 2 + b + 3 2 = a + 3 2 + b − 5 2 a + 3 2 + b − 5 2 = a − 3 2 + b − 5 2
⇔ a 2 − 6 a + 9 + b 2 + 6 b + 9 = a 2 + 6 a + 9 + b 2 − 10 b + 25 a 2 + 6 a + 9 + b 2 − 10 b + 25 = a 2 − 6 a + 9 + b 2 − 10 b + 25 ⇔ − 12 a + 16 b = 16 12 a = 0 ⇔ a = 0 b = 1
Vậy tâm I(0; 1).
Chọn B.