A=80 độ.

B=60 độ.

C=40 độ.

Xét tam giac ABC:  A + B + C=180 (*) 
mà B = 1,5.C => A + 1,5.C + C=180 
=> A = 180 - 2,5.C (1) 
Mặt khác: A1 = A2 = 90 - 1,25.C [từ (1)]
* Xét tam giác ADB ta có : 
A1 + B + góc BDA = 180 
=> 90 - 1,25.C + 1,5.C + 80 = 180 
=> C = 40 
thay C vào (1), ta có: 90 - 1,25.40 = 40 
nên A = 80 
từ (*) ta => B = 60 
Vậy A = 80, B = 60, C = 40

\(\widehat{ADB}=80^o\Rightarrow\widehat{ADC}=100^o\)

\(\widehat{B}+80^o+\frac{\widehat{A}}{2}=180^o\)

\(\widehat{C}+100^o+\frac{\widehat{A}}{2}=180^o\)

\(\Rightarrow\widehat{B}=\widehat{C}+20^o\)

\(\Rightarrow\frac{3}{2}\widehat{C}=\widehat{C}+20^o\)\(\Rightarrow\widehat{C}=40^o\Rightarrow\widehat{B}=60^o\Rightarrow\widehat{A}=80^o\)

tại sao từ 3/2 C = C-20maf => c=40độ luôn vậy bạn

Ta có : \(\widehat{D_1}+\widehat{D_2}=180^o\)

\(\Rightarrow\left\{{}\begin{matrix}\widehat{D_1}=80^o\\\widehat{D_2}=100^o\end{matrix}\right.\)

Mà \(\widehat{B}+\widehat{A_1}+\widehat{D_1}=\widehat{C}+\widehat{A_2}+\widehat{D_2}=180^o\)

\(\Leftrightarrow1,5\widehat{C}+80=\widehat{C}+100\)

\(\Rightarrow\widehat{C}=40^o\)

\(\Rightarrow\widehat{B}=60^o\)

\(\Rightarrow\widehat{A}=100-60-40=80^o\)

Vậy ...