Vì tứ giác ABCD có:

 \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\\ \Leftrightarrow\left(\widehat{A}+\widehat{B}\right)+50^o+80^o=360^o\\ \Leftrightarrow\widehat{A}+\widehat{B}=230^o\)

Mặt khác: \(\widehat{A}-\widehat{B}=20^o\Rightarrow\widehat{A}=20^o+\widehat{B}\)

\(\Rightarrow\widehat{B}+20^o+\widehat{B}=230^o\\ \Leftrightarrow2\widehat{B}+20^o=230^o\\ \Leftrightarrow2\widehat{B}=210^o\\ \Leftrightarrow\widehat{B}=210^o:2=105^o\\ \Rightarrow\widehat{A}=20^o+105^o=125^o\)

Tổng 4 góc trong tứ giác là 360^o 

⇒ \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}\)=360^o

⇒ \(\widehat{A}+\widehat{B}+\)50^o+80^o=360^o

⇒ \(\widehat{A}+\widehat{B}\)=230^o

\(\widehat{A}+\widehat{B}\)=230^o, \(\widehat{A}-\widehat{B}\)=20^o⇒\(\widehat{A}+\widehat{B}+\widehat{A}-\widehat{B}\)=250^o

                                                ⇒ \(2\widehat{A}\)=250^o

                                        ⇒ \(\widehat{A}\)=125^o

\(\widehat{A}+\widehat{B}\)=230^o

⇒ 125^o+\(\widehat{B}\)=230^o

⇒\(\widehat{B}\)=105^o

C

a: Đặt \(\dfrac{AB}{5}=\dfrac{AC}{12}=k\)

=>AB=5k; AC=12k

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(25k^2+144k^2=26^2\)

=>\(k^2=4\)

=>k=2

=>AB=10cm; AC=24cm

b: Xét tứ giác ABCD có

\(\widehat{A}+\widehat{B}+\widehat{BCD}+\widehat{ADC}=360^0\)

=>\(\widehat{BCD}+\widehat{ADC}=360^0-70^0=290^0\)

=>\(2\cdot\left(\widehat{ODC}+\widehat{OCD}\right)=290^0\)

=>\(\widehat{OCD}+\widehat{ODC}=145^0\)

Xét ΔOCD có \(\widehat{COD}+\widehat{OCD}+\widehat{ODC}=180^0\)

=>\(\widehat{COD}=180^0-145^0=35^0\)

Bài 1 :                                                   Bài giải

Ta có : \(\widehat{A}-\widehat{B}=10^o\text{ }\Rightarrow\text{ }\widehat{A}=\widehat{B}+10^o\)

Trong tứ giác ABCD có : 

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

\(\widehat{B}+10+\widehat{B}+60^o+80^o=360^o\)

\(2\widehat{B}+150^o=360^o\)

\(2\widehat{B}=110^o\)

\(\widehat{B}=55^o\text{ }\Rightarrow\text{ }\widehat{A}=65^o\)

Tổng bốn góc của 1 tứ giác bằng  360 °  nên: ∠A +  ∠ B +  ∠ C + ∠ D = 360 °

Suy ra:  ∠ A +  ∠ B =  360 °  – ( ∠ C + ∠ D) hay

∠ A +  ∠ B =  360 ° - 60 ° + 80 ° = 220 °  

Mà  ∠ A -  ∠ B = 10 °

Vậy  ∠ A =  = 115 ° ,  ∠ B =  115 °  -  10 °  =  105 °