các bn giúp mk nhé ai nhanh nhất mk tk cho.

1) 

Tổng của \(\widehat{B}\) và \(\widehat{C}\) là:

\(180^o-60^o=120^o\)

Ta có \(\widehat{B}=2\widehat{C}\Leftrightarrow\widehat{B}=\frac{2}{1}\widehat{C}\)

Áp dụng bài toán tổng tỉ.

Tổng số phần bằng nhau là:

2 + 1 = 3 phần.

Góc B là:

120 : 3 x 2 = 80 độ

Góc C là:

120 - 80 = 40 độ.

Vậy ......................

2) Theo đề ta có:

\(\frac{\widehat{A}}{2}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{4}\)  và \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{\widehat{A}}{2}=\frac{\widehat{B}}{3}=\frac{\widehat{C}}{4}=\frac{\widehat{A}+\widehat{B}+\widehat{C}}{2+3+4}=\frac{180^o}{9}=20^o\)

\(\hept{\begin{cases}\frac{\widehat{A}}{2}=20^o\Rightarrow\widehat{A}=20^o.2=40^o\\\frac{\widehat{B}}{3}=20^o\Rightarrow\widehat{B}=20^o.3=60^o\\\frac{\widehat{C}}{4}=20^o\Rightarrow\widehat{C}=20^o.4=80^o\end{cases}}\)

Vậy ..............................

a) ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{B}+\widehat{C}=100^0\Leftrightarrow\widehat{B}=100^0-\widehat{C}\)

mà \(\widehat{B}-\widehat{C}=20^0\Leftrightarrow100^0-\widehat{C}-\widehat{C}=20^0\Leftrightarrow\widehat{C}=40^0\)

vậy \(\widehat{B}=100^0-\widehat{C}=60^0\)

b) ta có \(\widehat{B}=3\widehat{C}\)

mà \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{B}+\widehat{C}=110^0\Leftrightarrow4\widehat{C}=110^0\Rightarrow\widehat{C}=27,5^0\)

\(\widehat{B}=3\widehat{C}=27,5^0.3=82,5^0\)

Xét \(\Delta ABC\)có : \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (tổng ba góc trong 1 tam giác)

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

<=> \(\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-75^o=105^o\)

Mà \(\widehat{B}=2\widehat{C}\)

Suy ra : \(2\widehat{C}+\widehat{C}=105^o\)

\(\Leftrightarrow3\widehat{C}=105^o\)

\(\Rightarrow\widehat{C}=\frac{105^o}{3}=35^o\)

\(\widehat{B}=105^o-35^o=70^o\)

Xét \(\Delta ABC\)có

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

Hay \(\widehat{B}+\widehat{C}=180^o-50^o\)

\(\widehat{B}+\widehat{C}=130^o\)

Suy ra :

    \(\widehat{B}=\frac{130^o+20^o}{2}=75^o\)

\(\widehat{C}=75^o-20^o=55^o\)

     Vậy \(\widehat{B}=75^o;\widehat{C}=55^o\)

Ta có : \(\widehat{B}+\widehat{C}=180^o-\widehat{A}=180^o-75^o=105^o\)

a/ \(\widehat{B}=2\widehat{C}\Rightarrow2\widehat{C}+\widehat{C}=105^o\Rightarrow3\widehat{C}=105^o\Rightarrow\widehat{C}=35^o\Rightarrow\widehat{B}=70^o\)

b/ \(\widehat{B}-\widehat{C}=25^o\Rightarrow\widehat{B}=\widehat{C}+25^o\Rightarrow\widehat{C}+25^o+\widehat{C}=105^o\Rightarrow2\widehat{C}=80^o\Rightarrow\widehat{C}=40^o\Rightarrow\widehat{B}=65^o\)

Bài 1: 

\(\widehat{A}\div\widehat{B}\div\widehat{C}=1\div2\div3=\frac{\widehat{A}}{1}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}\)

Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) (Tổng ba góc của một tam giác)

Áp dụng t/d dãy tỉ số bằng nhau, ta có: \(\frac{\widehat{A}}{1}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}=\frac{\widehat{A}+\widehat{B}+\widehat{C}}{1+2+3}=\frac{180^0}{6}=30\)

\(\Rightarrow\widehat{A}=30.1=30^0\)

     \(\widehat{B}=30.2=60^0\)

     \(\widehat{C}=30.3=90^0\)

Vậy .....

Bài 2: 

Gọi số đo các góc của tam giác ABC lần lượt là: a;b;c (\(a;b;c\inℕ^∗\) )

Ta có: \(a-b=18^0\Rightarrow a=18+b\)

          \(b-c=18^0\Rightarrow c=b-18\)

Trong tam giác ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

                      \(\Leftrightarrow a+b+c=180^0\)

                       \(\Leftrightarrow18+b+b+b-18=180^0\)

                        \(\Leftrightarrow3b=180^0\Rightarrow b=60\Rightarrow\widehat{B}=60^0\)

                          \(\Rightarrow\widehat{A}=18^0+\widehat{B}=18^0+60^0=78^0\)

                          \(\Rightarrow\widehat{C}=180^0-60^0-78^0=42^0\)

Vậy .....