`# \text {Ryo}`

`3,`

Xét tứ giác ABCD:

\(\widehat{\text{A}}+\widehat{\text{B}}+\widehat{\text{C}}+\widehat{\text{D}}=360^0\left(\text{định lý tổng các góc của tứ giác}\right)\\ \Rightarrow\widehat{x}+130^0+120^0+60^0=360^0\\ \Rightarrow\widehat{x}=50^0\)

Ta có:

`x + y = 180^0 (\text {kề bù})`

`=> 50^0 + y = 180^0`

`=> y = 130^0`

Vậy, `x = 50^0; y = 130^0`

Xét tứ giác MNJK:

\(\widehat{\text{M}}+\widehat{\text{N}}+\widehat{\text{J}}+\widehat{\text{K}}=360^0\\\Rightarrow \widehat{x}+90^0+90^0+90^0=360^0\\ \Rightarrow\widehat{x}=90^0\)

Vậy, `x = 90^0`

Xét tứ giác PQRS:

\(\widehat{\text{P}}+\widehat{\text{Q}}+\widehat{\text{R}}+\widehat{\text{S}}=360^0\\ \Rightarrow\widehat{\text{P}}+\widehat{\text{Q}}+95^0+65^0=360^0\\ \Rightarrow\widehat{\text{P}}+\widehat{\text{Q}}=200^0\)

Ta có:

\(\widehat{\text{P}}+\widehat{\text{Q}}=200^0\\ \Rightarrow x+3x=200^0\\ \Rightarrow4x=200^0\\ \Rightarrow x=50^0\)

Vì \(\widehat{\text{Q}}\) `= 3*x`

`=>`\(\widehat{\text{Q}}=3\cdot50^0=150^0\)

Vậy, `x = 50^0; 3x = 150^0`

Xét tứ giác MNPQ:

\(\widehat{\text{M}}+\widehat{\text{N}}+\widehat{\text{P}}+\widehat{\text{Q}}=360^0\\ \Rightarrow3x+4x+x+2x=360^0\\ \Rightarrow\left(3+4+1+2\right)x=360^0\\ \Rightarrow 10x=360^0\\ \Rightarrow x=36^0\)

Ta có:

\(\widehat{\text{M}}=3\cdot x\\ \widehat{\text{N}}=4\cdot x\\ \widehat{\text{Q}}=2\cdot x\)

\(\Rightarrow\widehat{\text{M}}=3\cdot36=108^0\\ \widehat{\text{N}}=4\cdot36=144^0\\ \widehat{\text{Q}}=2\cdot36=72^0\)

Vậy, `x = 36^0; 2x = 72^0; 3x = 108^0; 4x = 144^0.`

Ta có:

\(\widehat{\text{F}_2}+\widehat{\text{ EFG}}=180^0\left(\text{kề bù}\right)\\ \Rightarrow80^0+y=180^0\\ \Rightarrow y=100^0\)

Xét tứ giác EFGH:

\(\widehat{\text{E}}+\widehat{\text{F}}+\widehat{\text{G}}+\widehat{\text{H}}=360^0\\ \Rightarrow90^0+100^0+\text{ }\widehat{\text{G}}+\widehat{\text{H}}=360^0\\ \Rightarrow\widehat{\text{G}}+\widehat{\text{H}}=170^0\)

Ta có:

\(\widehat{\text{G}}+\widehat{\text{H}}=170^0\\ \Rightarrow x+x=170^0\\ \Rightarrow 2x=170^0\\ \Rightarrow x=85^0\)

Vậy, `x = 85^0; y = 100^0.`

Xét tứ giác ABDE:

\(\widehat{\text{A}}+\widehat{\text{B}}+\widehat{\text{D}}+\widehat{\text{E}}=360^0\\ \Rightarrow65^0+90^0+x+90^0=360^0\\ \Rightarrow x=115^0\)

Vậy, `x = 115^0.`

(1) \(x=360^o-60^o-130^o-120^o=50^o\)

\(y=180^o-50^o=130^o\)

(2) \(x=360^o-3\cdot90^o=90^o\)

(3) \(\widehat{Q}+\widehat{R}+\widehat{S}=65^o+95^o+3x=160^o+3x\)

\(x=360^o-160^o-3x\Leftrightarrow4x=200^o\Leftrightarrow x=50^o\)

(4) \(\widehat{M}+\widehat{N}+\widehat{Q}+\widehat{P}=360^o\)

\(\Rightarrow3x+4x+2x+x=360^o\)

\(\Leftrightarrow10x=360^o\)

\(\Leftrightarrow x=36^o\)

(5) \(y=180^o-80^o=100^o\)

\(\Rightarrow x+x=360^o-90^o-100^o\)

\(\Leftrightarrow2x=170^o\)

\(\Leftrightarrow x=85^o\)

(6) \(x=360^o-2\cdot90^o-65^o=115^o\)