kẻ xA//BC

\(=>\angle\left(A3\right)=\angle\left(C\right)\left(so-le-trong\right)\)

\(=>\angle\left(A1\right)=\angle\left(B\right)\left(so-le-trong\right)\)

mà \(\angle\left(A1\right)+\angle\left(A2\right)+\angle\left(A3\right)=180^o\left(ke-bu\right)\)

\(=>\angle\left(A2\right)+\angle\left(B\right)+\angle\left(C\right)=180^o\)

a)

 

=> Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180^o

100^o + \(\widehat{B}+\widehat{C}\) = 180^o

\(\widehat{B}+\widehat{C}\) = 180^o - 100^o

\(\widehat{B}+\widehat{C}\) = 80^o

Góc B = (80^o+50^o):2 = 65^o

=> \(\widehat{C}\) = 65^o - 50^o = 15^o

Vậy \(\widehat{B}\) = 65^o ; \(\widehat{C}\) = 15^o

b)

 

Ta có : \(\widehat{3A}+\widehat{B}+\widehat{2C}\) = 180^o

\(\widehat{3A}+\widehat{2C}\) = 180^o - 80^o

\(\widehat{3A}+\widehat{2C}\) = 100^o

=> \(\widehat{A}\) = 100^o:(3+2).3 = 60^o

\(\widehat{C}\) = 100^o - 60^o = 40^o

Vậy \(\widehat{A}\) = 60^o ; \(\widehat{C}\) = 40^o

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) 

Với ∆ABC thì các khẳng định

a) $\hat{A}+\hat{B}+\hat{C}>{180}^0$ là sai

b) $\hat{A}+\hat{B}<{180}^0$ là đúng

c)$\hat{B}+\hat{C}<{180}^0$ là đúng

d) $\hat{A}+\hat{B}\ge {180}^0$ là sai

b đúng
a, c, d sai

\(\dfrac{B}{C}=\dfrac{4}{3}\Rightarrow B=\dfrac{4C}{3}\)

\(B+C=180^0-A=105^0\Rightarrow C+\dfrac{4C}{3}=105^0\Rightarrow C=45^0\) \(\Rightarrow B=60^0\)

Kẻ đường cao AD ứng với BC (do 2 góc B và C đều nhọn nên D nằm giữa B và C)

Trong tam giác vuông ABD:

\(sinB=\dfrac{AD}{AB}\Rightarrow AD=AB.sinB=10,6.sin60^0\approx9,2\left(cm\right)\)

\(cosB=\dfrac{BD}{AB}\Rightarrow BD=AB.cosB=10,6.cos60^0=5,3\left(cm\right)\)

Trong tam giác vuông ACD:

\(tanC=\dfrac{AD}{CD}\Rightarrow CD=AD.tanC=9,2.tan45^0=9,2\left(cm\right)\)

\(sinC=\dfrac{AD}{AC}\Rightarrow AC=\dfrac{AD}{sinC}=\dfrac{9,2}{sin45^0}\approx13\left(cm\right)\)

\(BC=BD+CD=5,3+9,2=14,5\left(cm\right)\)

\(S_{ABC}=\dfrac{1}{2}AD.BC=\dfrac{1}{2}.9,2.14,5=66,7\left(cm^2\right)\)

\(\widehat{B}+\widehat{C}=140^0\)

\(\Leftrightarrow4\cdot\widehat{C}=140^0\)

\(\Leftrightarrow\widehat{C}=35^0\)

hay \(\widehat{B}=105^0\)

Vậy:  ΔABC tù