Cho tam giác ABC có A = 70%, B – C = 400. Tính số đo góc B, góc C của tam giác. Tam giác đã cho là tam giác vuông, nhọn hay từ?
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a)A=70
b)tam giac ABC la tam giac nhon vi co so do 3 canh <90
tuwj vex hinhf nha
1 a. xét tam giác abc có
góc a + góc b + góc c = 180 độ
t/s vào tính đc góc b + góc c= 120 độ
góc acb = 120 độ : ( 2+1).1=40 độ
b) xét tam giác abc có
góc a + góc b + góc c = 180 độ
t/s vào tính đc góc abc = 80 độ
có bi là tia phân giác của góc abc
=> góc abi = góc ibc = 80 độ :2=40 độ
có ci là tia phân giác của góc acb
=> góc aci = gócicb = 40 độ : 2 = 20 độ
xét tam giác ibc có
góc bic + góc ibc + bci = 180độ
thay số vào tính đc góc bic = 120 đọ( nghĩ z chứ chưa tính kĩ nha )
2
Ta có: ΔABD vuông cân tại B(gt)
nên \(\widehat{DAB}=45^0\)(Số đo của một góc nhọn trong ΔABD vuông cân tại B)
Ta có: ΔACE vuông cân tại C(gt)
nên \(\widehat{EAC}=45^0\)(Số đo của một góc nhọn trong ΔACE vuông cân tại C)
Ta có: ΔABC đều(gt)
nên AB=AC=BC và \(\widehat{BAC}=60^0\)(Số đo của các cạnh và các góc trong ΔABC đều)(1)
Ta có: \(\widehat{DAE}=\widehat{DAB}+\widehat{BAC}+\widehat{EAC}\)
\(\Leftrightarrow\widehat{DAE}=60^0+45^0+45^0=150^0\)
Ta có: ΔADB vuông cân tại B(gt)
nên AB=BD(hai cạnh bên)(2)
Ta có: ΔACE vuông cân tại C(gt)
nên AC=CE(hai cạnh bên)(3)
Từ (1), (2) và (3) suy ra AB=BC=AC=CE=DB
Xét ΔABD vuông tại B và ΔACE vuông tại C có
AB=AC(cmt)
DB=EC(cmt)
Do đó: ΔABD=ΔACE(hai cạnh góc vuông)
hay AD=AE(hai cạnh tương ứng)
Xét ΔADE có AD=AE(cmt)
nên ΔADE cân tại A(Định nghĩa tam giác cân)
hay \(\widehat{ADE}=\widehat{AED}=\dfrac{180^0-\widehat{DAE}}{2}\)(Số đo của các góc ở đáy trong ΔADE cân tại A)
hay \(\widehat{ADE}=15^0\) và \(\widehat{AED}=15^0\)
Vậy: Số đo các góc nhọn trong ΔADE là 150
Xét tam giác ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)( tổng 3 góc trong tam giác)
\(\Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=180^0-70^0=110^0\)
Xét tam giác ABC có:
\(\left\{{}\begin{matrix}\widehat{B}+\widehat{C}=110^0\\\widehat{B}-\widehat{C}=40^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(110^0+40^0\right):2=75^0\\\widehat{C}=\left(110^0-40^0\right):2=35^0\end{matrix}\right.\)
Ta có: \(\widehat{C}< \widehat{A}< \widehat{B}< 90^0\)
Vậy tam giác ABC là tam giác nhọn
OMG iu cậu^^