cho tam giác ABC có góc B;C nhọn.qua B kẻ BD vuông góc với AC.qua C kẻ CE vuông góc với AB.gọi H là giao điểm của BD và CE hãy tìm mối liên hệ
a/góc ABD và góc ACE
b/ góc A và góc DHE
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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)
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
NL
Nguyễn Lê Phước Thịnh
CTVHS
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
NL
Nguyễn Lê Phước Thịnh
CTVHS
17 tháng 8 2023
3:
góc C=90-50=40 độ
Xét ΔABC vuông tại A có sin C=AB/BC
=>4/BC=sin40
=>\(BC\simeq6,22\left(cm\right)\)
\(AC=\sqrt{BC^2-AB^2}\simeq4,76\left(cm\right)\)
1:
góc C=90-60=30 độ
Xét ΔABC vuông tại A có
sin B=AC/BC
=>3/BC=sin60
=>\(BC=\dfrac{3}{sin60}=2\sqrt{3}\left(cm\right)\)
=>\(AB=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\left(cm\right)\)
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`
22 tháng 2 2020
Câu hỏi của Nguyễn Vũ Thu Hương - Toán lớp 7 - Học toán với OnlineMath
NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 4
câu 5: Gọi M là giao điểm của AD và BC
Xét ΔBAD có \(\hat{BDM}\) là góc ngoài tại đỉnh D
nên \(\hat{BDM}=\hat{DAB}+\hat{DBA}\)
=>\(\hat{BDM}>\hat{BAD}=\hat{BAM}\) (2)
Xét ΔDAC có \(\hat{MDC}\) là góc ngoài tại đỉnh D
nên \(\hat{MDC}=\hat{DAC}+\hat{DCA}>\hat{DAC}\) (1)
Từ (1),(2) suy ra \(\hat{BDM}+\hat{MDC}>\hat{BAD}+\hat{CAD}\)
=>\(\hat{BDC}>\hat{BAC}\)
Câu 3:
Theo đề, ta có: \(\hat{A}=\hat{B}+25^0;\hat{C}=\hat{B}+35^0\)
Xét ΔBAC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{B}+\hat{B}+25^0+\hat{B}+35^0=180^0\)
=>\(3\cdot\hat{B}=180^0-60^0=120^0\)
=>\(\hat{B}=\frac{120^0}{3}=40^0\)
=>\(\hat{C}=40^0+35^0=75^0\)
Bài 2:
Theo đề, ta có: \(\hat{B}=\hat{A}+24^0;\hat{C}=\hat{A}-30^0\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{A}+\hat{A}+24^0+\hat{A}-30^0=180^0\)
=>\(3\cdot\hat{A}=180^0+30^0-24^0=186^0\)
=>\(\hat{A}=62^0\)
=>\(\hat{C}=62^0-30^0=32^0\)
Câu 1: Theo đề, ta có: \(\hat{B}=\hat{A}+15^0;\hat{C}=\hat{A}+45^0\)
Xét ΔABC có \(\hat{A}+\hat{B}+\hat{C}=180^0\)
=>\(\hat{A}+\hat{A}+15^0+\hat{A}+45^0=180^0\)
=>\(3\cdot\hat{A}=180^0-60^0=120^0\)
=>\(\hat{A}=40^0\)
\(\hat{B}=40^0+15^0=55^0\)