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Ta có : \(\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-75^o=105^o\)
a/ \(\widehat{B}=2\widehat{C}\Rightarrow2\widehat{C}+\widehat{C}=105^o\Rightarrow3\widehat{C}=105^o\Rightarrow\widehat{C}=35^o\Rightarrow\widehat{B}=70^o\)
b/ \(\widehat{B}-\widehat{C}=25^o\Rightarrow\widehat{B}=\widehat{C}+25^o\Rightarrow\widehat{C}+25^o+\widehat{C}=105^o\Rightarrow2\widehat{C}=80^o\Rightarrow\widehat{C}=40^o\Rightarrow\widehat{B}=65^o\)
A B C 110*
=> \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)
\(\widehat{A}+\widehat{B}=180^o-110^o\)
\(\widehat{A}+\widehat{B}=70^o\)
=> \(\widehat{A}\) = 70o:(3+4).3 = 30o
=> \(\widehat{B}\) = 70o - 30o = 40o
Vậy  = 30o ; \(\widehat{B}\) = 40o và \(\widehat{C}\) = 110o
a)
A B C 100*
=> Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180o
100o + \(\widehat{B}+\widehat{C}\) = 180o
\(\widehat{B}+\widehat{C}\) = 180o - 100o
\(\widehat{B}+\widehat{C}\) = 80o
Góc B = (80o+50o):2 = 65o
=> \(\widehat{C}\) = 65o - 50o = 15o
Vậy \(\widehat{B}\) = 65o ; \(\widehat{C}\) = 15o
b)
80* A B C
Ta có : \(\widehat{3A}+\widehat{B}+\widehat{2C}\) = 180o
\(\widehat{3A}+\widehat{2C}\) = 180o - 80o
\(\widehat{3A}+\widehat{2C}\) = 100o
=> \(\widehat{A}\) = 100o:(3+2).3 = 60o
\(\widehat{C}\) = 100o - 60o = 40o
Vậy \(\widehat{A}\) = 60o ; \(\widehat{C}\) = 40o
a) góc A = 70o, => B + C = 110o
=> B =(110 + 10) : 2 = 60
C = 60 - 10 = 50
b) góc A = 100 , => B + C = 80
=> B = (80 + 50) : 2 = 65
C = 65 - 50 = 15
c) B = 2C => 180 - 60 = 3C = 120
=> C = 40
=> B = 40 . 2 = 80
a) ΔABC có:
\(\widehat{A}\) + \(\widehat{B}\) + \(\widehat{C}\) = 180o hay 100o + \(\widehat{B}\) + \(\widehat{C}\) = 180o
\(\Rightarrow\) \(\widehat{B}\) + \(\widehat{C}\) = 180o - 100o = 80o
Ta có: \(\widehat{B}\) + \(\widehat{C}\) = 80o(cm trên) ; \(\widehat{B}\) - \(\widehat{C}\) = 50o (gt)
\(\Rightarrow\) \(\widehat{B}\) = (80o + 50o ) : 2 = 65o
\(\widehat{C}\) = (80o - 50o) : 2 = 15o
b) ΔABC có:
\(\widehat{B}\) + \(\widehat{A}\) + \(\widehat{C}\) = 180o hay 80o + \(\widehat{A}\) + \(\widehat{C}\) = 180o
\(\Rightarrow\) \(\widehat{A}\) + \(\widehat{C}\) = 180o - 80o = 100o
Ta có: 3 . \(\widehat{A}\) = 2 . \(\widehat{C}\) => \(\frac{\widehat{A}}{2}\) = \(\frac{\widehat{C}}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{\widehat{A}}{2}\) = \(\frac{\widehat{C}}{3}\) = \(\frac{\widehat{A}+\widehat{C}}{2+3}\) = \(\frac{100}{5}\) = 20
\(\Rightarrow\) \(\begin{cases}\widehat{A}=40^o\\\widehat{C}=60^o\end{cases}\)
a, \(3\widehat{A}=4\widehat{B}\Leftrightarrow\dfrac{3\widehat{A}}{12}=\dfrac{4\widehat{B}}{12}\Rightarrow\dfrac{\widehat{A}}{4}=\dfrac{\widehat{B}}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{\widehat{A}}{4}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{A}-\widehat{B}}{4-3}=\dfrac{20^0}{1}=20^0\)
+)\(\dfrac{\widehat{A}}{4}=20^0\Rightarrow\widehat{A}=20^0.4=80^0\)
+)\(\dfrac{\widehat{B}}{3}=20^0\Rightarrow\widehat{B}=20^0.3=60^0\)
Xét △ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ 80^0+60^0+\widehat{C}=180^0\\ \widehat{C}=180^0-80^0-60^0=40^0\)
Vậy \(\Delta ABC\) có \(\widehat{A}=80^0;\widehat{B}=60^0;\widehat{C}=40^0\)
a) Gọi số đo các góc lần lượt là x,y ( x,y > 0 )
Theo bài ra ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}\) và \(x-y=20^0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x-y}{4-3}=\dfrac{20^0}{1}=20^0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=20^0\Rightarrow x=80^0\\\dfrac{y}{3}=20^0\Rightarrow x=60^0\end{matrix}\right.\)
Xét \(\Delta ABC\) có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
mà \(\widehat{A}=80^0;\widehat{B}=60^0\)
\(\Rightarrow80^0+60^0+\widehat{C}=180^0\)
\(\Rightarrow140^0+\widehat{C}=180^0\)
\(\Rightarrow\widehat{C}=180^0-140^0\)
\(\Rightarrow\widehat{C}=40^0\)
Vậy ........................
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