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a)
=> 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)
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
Xét ΔABC có
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
\(\Leftrightarrow2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-\alpha\)
\(\Leftrightarrow\widehat{IBC}+\widehat{ICB}=\dfrac{180^0-\alpha}{2}\)
Xét ΔIBC có
\(\widehat{BTC}+\widehat{IBC}+\widehat{ICB}=180^0\)
\(\Leftrightarrow\widehat{BTC}=180^0-\dfrac{180^0-\alpha}{2}=\dfrac{180^0+\alpha}{2}\)
Ta có \(\widehat{A}+\widehat{ABC}+\widehat{C}=180^0\Rightarrow180^0-3\widehat{C}+\widehat{C}=180^0-70^0=110^0\)
\(\Rightarrow2\widehat{C}=70^0\Rightarrow\widehat{C}=35^0\Rightarrow\widehat{A}=180^0-3\cdot35^0=75^0\)
Ta có BE là p/g nên \(\widehat{B_1}=\widehat{B_2}=\dfrac{1}{2}\widehat{ABC}=35^0\)
Mà \(ED//BC\) nên \(\widehat{B_2}=\widehat{E_2}=35^0\left(so.le.trong\right)\left(1\right)\)
Ta có \(ED//BC\Rightarrow\widehat{E_1}=\widehat{C}=35^0\left(đồng.vị\right)\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\widehat{E_1}=\widehat{E_2}\left(=35^0\right)\)
Vậy ...
Tổng ba góc trong một tam giác bằng 180°. Vậy trong tam giác A’B’C’ có \(\widehat {C'} = 180^\circ - 70^\circ - 60^\circ = 50^\circ \).
Xét hai tam giác ABC và A’B’C’ có:
\(\widehat B = \widehat {B'} = 60^\circ ;\)
BC = B’C’ ( = 3 cm)
\(\widehat C = \widehat {C'} = 50^\circ \)
Vậy \(\Delta ABC = \Delta A'B'C'\)(g.c.g)
\(\widehat{B}+\widehat{C}=140^0\)
\(\Leftrightarrow4\cdot\widehat{C}=140^0\)
\(\Leftrightarrow\widehat{C}=35^0\)
hay \(\widehat{B}=105^0\)
Vậy: ΔABC tù
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Rightarrow180^0-3\widehat{C}+\widehat{B}+\widehat{C}=180^0\Rightarrow-2\widehat{C}+\widehat{B}=0^0\Rightarrow\widehat{B}=2\widehat{C}\)