cho tam giác ABC có góc B; góc C nhọn, đường cao AH
a) Chứng minh: AH=\(\dfrac{BC}{cotB+cotC}\)
b) Tính SABC, biết BC=4cm; góc B=450; góc C=300
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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)
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
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
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
=>\(BC=\dfrac{3}{sin60}=2\sqrt{3}\left(cm\right)\)
=>\(AB=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\left(cm\right)\)
`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`
Câu hỏi của Nguyễn Vũ Thu Hương - Toán lớp 7 - Học toán với OnlineMath
a.
Trong tam giác vuông ABH ta có:
\(cotB=\dfrac{BH}{AH}\Rightarrow BH=AH.cotB\)
Trong tam giác vuông ACH ta có:
\(cotC=\dfrac{CH}{AH}\Rightarrow CH=AH.cotC\)
\(\Rightarrow BH+CH=AH.cotB+AH.cotC\)
\(\Leftrightarrow BC=AH\left(cotB+cotC\right)\)
\(\Leftrightarrow AH=\dfrac{BC}{cotB+cotC}\) (đpcm)
b. Áp dụng công thức câu a:
\(AH=\dfrac{4}{cot45^0+cot30^0}=-2+2\sqrt{3}\) (cm)
\(S_{ABC}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}.\left(-2+2\sqrt{3}\right).4=-4+4\sqrt{3}\approx2,93\left(cm^2\right)\)