`#Ly`

$a)$ Ta có:

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

$⇒$ $\widehat{D}$ $=$ $360^o$ $-$ $\widehat{A}$ $-$ $\widehat{B}$ $-$ $\widehat{C}$

$=$ $360^o$ $-$ $50^o$ $-$ $100^o$ $-$ $120^o$ $=$ $90^o$

$b)$ Xét tứ giác $MNPQ$ có:

$\widehat{M}$ $=$ $\widehat{N}$ $=$ $\widehat{Q}$ $=$ $90^o$

$⇒$ $MNPQ$ là hình chữ nhật

$⇒$ $\widehat{N}$ $=$ $90^o$

$c)$ Ta có:

$\widehat{F}$ $+$ $\widehat{E}$ $+$ $\widehat{H}$ $+$ $\widehat{G}$ $=$ $360^o$

$⇒$ $\widehat{G}$ $=$ $360^o$ $-$ $\widehat{F}$ $-$ $\widehat{E}$ $-$ $\widehat{H}$

$=$ $360^o$ $-$ $90^o$ $-$ $100^o$ $-$ $90^o$ $=$ $80^o$

$d)$ Ta có:

$\widehat{IKL}$ $=$ $180^o-100^o=80^o$

$\widehat{KLR}$ $=$ $180^o-60^o=120^o$

$\widehat{IKL}$ $+$ $\widehat{I}$ $+$ $\widehat{R}$ $+$ $\widehat{KLR}$ $=$ $360^o$

$⇒$ $\widehat{I}$ $=$ $360^o$ $-$ $\widehat{IKL}$ $-$ $\widehat{R}$ $-$ $\widehat{KLR}$

$=$ $360^o$ $-$ $80^o$ $-$ $90^o$ $-$ $120^o$ $=$ $70^o$

Đáp án:

Ta có: `hat(K_1) + hat(IKL) = 180^0` `(2` góc kề bù `)`

Thay số: `100^0 + hat(IKL) = 180^0`

`=>` `hat(IKL) = 80^0`

Ta có: `hat(L_1) + hat(KLR) = 180^0` `(2` góc kề bù `)`

Thay số: `60^0 +hat(KLR) = 180^0`

`=>` `hat(KLR) = 120^0`

Vì `LR ⊥ RI` (hình vẽ)

`=>` `hat(LRI) = 90^0`

Xét tứ giác `KLRI` có:

`hat(IKL) + hat(KLR) + hat(LRI) + hat(I) = 360^0`

Thay số: `80^0 + 120^0 + 90^0 + x = 360^0`

`=> 290^0 + x = 360^0`

`=> x = 360^0 - 290^0`

`=> x = 70^0`

Vậy `x = 70^0`

`#dariana`