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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:31

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Tiến Phạm
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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

=>\(\hat{B}=45^0\)

=>\(\hat{A}+\hat{C}=180^0-45^0=135^0\)

\(\hat{A}-\hat{C}=-5^0\)

nên \(\hat{A}=\frac{135^0-5^0}{2}=\frac{130^0}{2}=65^0\)

=>\(\hat{C}=135^0-65^0=70^0\)

Đỗ Vũ Nhật Anh
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Duy Nghĩa Hoàng
15 tháng 11 2021 lúc 21:58

Giống mình làm

 

Bà HOÀng Thả ThÍnh
Xem chi tiết
Dương Mạnh Quyết
21 tháng 12 2021 lúc 10:21

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)

Khách vãng lai đã xóa
Lưu Nguyễn Hà An
15 tháng 2 2022 lúc 9:04

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

Trần Thị Thu Mến
31 tháng 10 2024 lúc 18:47

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)

Nguyễn thư
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Nguyễn thị ngọc hoan
Xem chi tiết
Minh Nhân
23 tháng 1 2021 lúc 15:41

#TK

Suy ra diện tích tam giác ABC là 1/2.AB.BC = 6.

 
Pham Trong Bach
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Cao Minh Tâm
11 tháng 10 2017 lúc 18:29

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.

Khánh Linh Bùi
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Phương Thảo
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Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:29

chịu hoi =))))))

 

Trịnh Việt Dũng
15 tháng 6 2022 lúc 20:29

em mới học lớp 7 hà

năm nay lên lớp 8 =)))))

Nguyễn Thảo My
14 tháng 1 2023 lúc 21:25

1)Ta có: \(S_{ABC}=\dfrac{1}{2}AB.AC.\sin A\)

\(\Leftrightarrow8=\dfrac{1}{2}\times4\times5\times sinA\)

\(\Leftrightarrow\sin A=0,8\)

Lại có: \(\left(\sin A\right)^2+\left(\cos A\right)^2=1\Leftrightarrow\cos A=0,6.\)

Áp dụng định lí hàm số cosin:

\(BC^2=AB^2+AC^2-2AB\times AC\times\cos A\)

\(\Leftrightarrow BC^2=4^2+5^2-2\times4\times5\times0,6=17\)

\(\Leftrightarrow BC=\sqrt{17}.\)

2) Trong \(\Delta ABC\) có: \(g\text{ó}cA+g\text{óc}B+g\text{óc}C=180^o\)

=> BAC=75o.

Áp dụng định lí hàm số sin:

\(\dfrac{AB}{\sin C}=\dfrac{BC}{\sin A}\Leftrightarrow\dfrac{3}{\sin45^o}=\dfrac{BC}{\sin75^o}\)

\(\Leftrightarrow BC=\dfrac{3+3\sqrt{3}}{2}\).

 

 

Pham Trong Bach
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Cao Minh Tâm
29 tháng 6 2018 lúc 7:22

Chọn B.

Ta có:

Mặt khác 

Suy ra diện tích tam giác ABC là 1/2.AB.BC = 6.