Cho tam giác ABC có góc B, C nhọn; M nằm trên BC. Gọi \(d\) là tổng của khoảng cách từ B đến AM và khoảng cách từ C đến AM. Tìm M để \(d\) có giá trị max
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15 tháng 6 2022
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24 tháng 8 2021
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
17 tháng 9 2023
a) Tam giác ABC nhọn:
b) Tam giác ABC vuông tại A:
c) Tam giác ABC có góc A tù:
CC
15 tháng 12 2019
\(\Delta ABC\)có \(\widehat{A}=60^o\)\(\Rightarrow\widehat{B}+\widehat{C}=120^o\)
mà \(\widehat{B}=3.\widehat{C}\)\(\Rightarrow4.\widehat{C}=120^o\)\(\Rightarrow\widehat{C}=30^o\)\(\Rightarrow\widehat{B}=90^o\)
\(\Rightarrow\Delta ABC\)vuông tại B
PT
1
PT
1
CM
25 tháng 3 2017
Do B tù nên ta có góc ngoài của đỉnh B là góc nhọn, suy ra các góc A, C nhọn.
NP
2
29 tháng 7 2017
vì góc C tù nên góc C lớn hơn 90 độ
Nên góc A, B chỉ có thể là góc nhọn, không thể là góc vuông vì một tam giác có tổng là 180 độ
kbn nha
29 tháng 7 2017
Tam giác ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180\)
Mà \(\widehat{C}\)> 90
\(\Rightarrow\widehat{A}+\widehat{B}< 90\)
\(\Rightarrow\widehat{A}< 90;\widehat{B}< 90\)
14 tháng 6 2017
Câu 1:
Xét tam giác ABH vuông tại H, ta có:
AB2 = AH2 + HB2 (định lý Py-ta-go)
202 = AH2 + 162
400 = AH2 + 256
AH2 = 400 - 256
AH2 = 144
AH = \(\sqrt{144}\)= 12 (cm)
Xét tam giác AHC vuông tại H, ta có:
AC2 = AH2 + HC2 (định lý Py-ta-go)
AC2 = 122 + 52
AC2 = 144 + 25
AC2 = 169
AC = \(\sqrt{169}\)= 13 (cm)
Vậy AH = 12 cm
AC = 13 cm
Bài 2:
Xét tam giác AHC vuông tại H, ta có:
AC2 = AH2 + HC2 (định lý Py-ta-go)
152 = AH2 + 92
225 = AH2 + 81
AH2 = 225 - 81
AH2 = 144
AH = \(\sqrt{144}\)= 12 (cm)
Xét tam giác AHB vuông tại, ta có:
AB2 = AH2 + HB2 (định lý Py-ta-go)
AB2 = 122 + 52
AB2 = 144 + 25
AB2 = 169
AB = \(\sqrt{169}\)= 13 (cm)
Vậy AB = 13 cm
khoảng cách là vuông góc bạn nhé
đặt 2 điểm o và k lần lượt là điểm mà b vuông góc vs am và c vuông góc vs am
suy ra
bo < bm đường xiên và đường vuông góc
ck < cm DXVDVG giống trên nha bạn
suy ra d lớn nhất bằng bm + cm = cb
suy ra bm vuông góc vs am
vậy am là đường cao