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

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

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

hay \(\widehat{B}=105^0\)

Vậy:  ΔABC tù

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

Do ΔABC cân tại B => A = C = \(\dfrac{180^o-80^o}{2}=50^o\)

=> góc BAI = 50^o - 10^o = 40^o 

góc BCI = 50^o - 30^o = 20^o

=> \(IBC=\dfrac{1}{3}ABI\Rightarrow IBC=\dfrac{80^o}{3+1}=20^o;ABI=80^o-20^o=60^o\)

\(\Leftrightarrow AIB=180^o-40^o-60^o=80^o\)

a. Ta thấy \(\left(a\sqrt{5}\right)^2=\left(a\sqrt{3}\right)^2+\left(a\sqrt{2}\right)^2\Rightarrow AB^2=BC^2+AC^2\)

\(\Rightarrow\Delta ABC\)vuông tại C

b. \(\sin B=\frac{AC}{AB}=\frac{\sqrt{2}}{\sqrt{5}}=\frac{\sqrt{10}}{5};\cos B=\frac{CB}{AB}=\frac{\sqrt{3}}{\sqrt{5}}=\frac{\sqrt{15}}{5}\)

\(\tan B=\frac{AC}{AB}=\frac{\sqrt{6}}{3};\cot B=\frac{\sqrt{6}}{2}\)

\(\sin A=\cos B=\frac{\sqrt{15}}{5};\cos A=\sin B=\frac{\sqrt{10}}{5}\)

\(\tan A=\cot B=\frac{\sqrt{6}}{2};\cot A=\tan B=\frac{\sqrt{6}}{3}\) 

Thanks bạn nhìu

Xét \(\Delta ABC\) và \(\Delta HBA\) có:

\(\widehat{BAC}=\widehat{BHA}=90^o\)

\(\widehat{B}\) chung

\(\Rightarrow\Delta ABC\sim\Delta HBA\left(g.g\right)\) (1)

\(\Rightarrow\dfrac{AB}{BC}=\dfrac{HB}{AB}\) hay \(\dfrac{AB}{4+9}=\dfrac{4}{AB}\Rightarrow AB^2=52\Rightarrow AB=\sqrt{52}=2\sqrt{13}cm\)

Xét \(\Delta\text{A}BC\) và \(\Delta HAC\) có:

\(\widehat{BAC}=\widehat{AHC}=90^o\)

\(\widehat{C}\) chung

\(\Rightarrow\Delta ABC\sim\Delta HAC\left(g.g\right)\) (2)

Từ (1) và (2) \(\Rightarrow\Delta HAB\sim\Delta HCA\)

\(\Rightarrow\dfrac{AH}{HC}=\dfrac{HB}{AH}\) hay \(\dfrac{AH}{9}=\dfrac{4}{AH}\Rightarrow AH^2=36\Rightarrow AH=\sqrt{36}=6\left(cm\right)\)

Ta có \(\Delta ABC\) vuông tại A.

Áp dụng đinh lý Py-ta-go ta có:

\(AC=\sqrt{BC^2-AB^2}=\sqrt{\left(4+9\right)^2-\left(2\sqrt{13}\right)^2}=3\sqrt{13}cm\)

b) Diện tích tam giác ABC là:

\(S_{ABC}=\dfrac{1}{2}\cdot BC\cdot AH=\dfrac{1}{2}\cdot\left(4+9\right)\cdot6=39\left(cm^2\right)\)

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Xét tam giác ABC ta có \(\widehat{ADB}\)=80 độ

\(\widehat{ABC}\)= 80 độ => Góc BAD = 20 độ. => góc DAC =20 độ

=> Góc C = 60 độ 

Vậy Góc A= 40 độ; B= 80 độ ; C= 60độ

Giải
Xét tgiac ABD có: \(\widehat{B}=\widehat{ADB}=80^0\)\(\Rightarrow\widehat{BAD}=180^0-\left(\widehat{B}+\widehat{ADB}\right)\)( Áp dụng tính chất tổng ba góc trong tam giác)
\(=180^0-\left(80^0+80^0\right)=20^0\)
Lại có: \(\widehat{BAD}=\widehat{DAC}\)( vì AD là tia phan giác góc A) nên:
\(\Rightarrow\widehat{BAD}=\widehat{DAC}=20^0\Rightarrow\widehat{BAC}=20^0\times2=40^0\)
\(\Rightarrow\widehat{ACB}=180^0-\left(\widehat{BAC}+\widehat{ABC}\right)=180^0-\left(40^0+80^0\right)\)\(=60^0\)
Vậy, \(\widehat{A}=40^0;\widehat{B}=80^0;\widehat{C}=60^0\)