Cho tam giác ABC có góc ABC<90o.Trên nửa mặt phẳng bờ là đường thẳng BC có chứa điểm A.Vẽ tia Bx vuông góc với BC.Trên tia Bx lấy D sao cho BD=BC.Trên nửa mặt phẳng là đường thẳng AB có chứa điểm C,vẽ By vuông góc với BA.Trên tia By lấy E sao cho BE=BA.CMR:a/DA=D,b/\(DA\perp CE.\)
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15 tháng 6 2022
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21 tháng 12 2021
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)
LN
Lưu Nguyễn Hà An
CTVHS
15 tháng 2 2022
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
23 tháng 12 2021
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{1}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{a+b+c}{1+3+5}=\dfrac{180}{9}=20\)
Do đó: a=20; b=60; c=100
Vậy: ΔABC là tam giác tù
YN
12 tháng 5 2021
* Theo mình thì phần a) Góc A = 90 độ sẽ hợp lý hơn chứ. Vậy nên mình sẽ làm theo cả hai góc A 90 độ và 80 độ nhé ( Nhưng bài của mình phần b) sẽ theo góc A = 90 độ )
a)
Góc A = 80 độ thì sẽ có thể tam giác ABC là tam giác cân, tam giác ⊥ tại B hoặc C, tam giác ABC là tam giác tù hoặc tam giác nhọn
Góc A = 90 độ thì tam giác ABC là tam giác vuông tại A
b)
Theo phần a), ta có: Tam giác ABC cân tại A
=> Góc B = góc C = ( 180 độ - 70 độ ) : 2 = 55 độ
19 tháng 1 2023
`a,` vì Tam giác `ABC` có \(\widehat{A}=110^0\)
`=>` Tam giác `ABC` là tam giác tù.
`b,` Cạnh đối diện của \(\widehat{A}\) là cạnh `BC`
`=>` Cạnh lớn nhất của Tam giác `ABC` là cạnh `BC`
