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

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 ...

Ta có : \(\widehat{A_2}=180^o-\widehat{B}-\widehat{D_2}\Rightarrow\widehat{A_2}=100^o-\widehat{B}\)

\(\widehat{A_1}=180^o-\widehat{D_1}-\widehat{C}\)

\(=80^o-\widehat{C}\)

\(\Rightarrow100^o+\widehat{B}=80^o-\widehat{C}\)

\(\Rightarrow100^o-1,5.\widehat{C}=80^o-\widehat{C}\)

\(\Rightarrow100^o-80^o=-\widehat{C}+1.5.\widehat{C}\)

\(\Rightarrow20^o=\dfrac{1}{2}.\widehat{C}\Rightarrow\widehat{C}=40^o\)

Thay \(\widehat{C}=40^o\) vào \(\widehat{A}_1\), ta có:

\(\widehat{A_1}=80^o-40^o=40^o\)

Ta có \(\widehat{A_1}=\widehat{A_2}\Rightarrow\widehat{A}=40^o.2=80^o\)

\(\Rightarrow\widehat{B}=180^o-\left(\widehat{A}+\widehat{C}\right)\)

\(=180^o-\left(80^o+40^o\right)=60^o\)

Vậy \(\left\{{}\begin{matrix}\widehat{A}=80^o\\\widehat{B}=60^o\\\widehat{C}=40^o\end{matrix}\right.\)

~ Học tốt ~