Câu 34: `C. 55^o`

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Xét `ΔABC` có:

$\widehat{BAC}+\widehat{ABC}+\widehat{ACB}=180^o$     ( Tổng `3` góc trong tam giác)

`->`$\widehat{BAC}=180^o-100^o-30^o=50^o$

`*`

`BD` là phân giác góc `A`

`->`$\widehat{BAD}=\widehat{DAC}=\dfrac{\widehat{BAC}}{2}=25^o$

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Xét `ΔABD` có:

$\widehat{ABD}+\widehat{BAD}+\widehat{ADB}=180^o$  ( Tổng `3` góc trong tam giác)

`->`$\widehat{ADB}=180^o-100^o-25^0=55^o $

Câu 35:  `D. 75^o`

`*``MN////BC`

`->`$\widehat{MBC}=\widehat{AMN}=55^o$   (Góc đồng vị)

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$\widehat{ANM}+\widehat{MNC}=180^o$   ( Kề bù)

`->`$\widehat{ANM}=180^o-130^o=50^o$

`*`

Xét `ΔAMN` có:

$\widehat{MAN}+\widehat{AMN}+\widehat{ANM}=180^o$   ( Tổng `3` góc trong tam giác)

`->`$\widehat{MAN}=180^o-50^o-55^o=75^o $

Câu 34 $\textit{A}$

$\longrightarrow$ $\Delta$ABC=$\widehat{A}$+$\widehat{B}$+$\widehat{C}$=$180^0$

$\longrightarrow$ $\widehat{A}$=$180^0$-$100^0$-$30^0$=$50^0$

$\longrightarrow$ $\widehat{ADB}$=$50^0$-$25^0$=$250$

Câu35 $\textit{D}$

$\longrightarrow$ $\widehat{C}$=180-130=50

$\longrightarrow$  $\Delta$ABC=$\widehat{A}$+$\widehat{B}$+$\widehat{C}$=$180^0$

$\longrightarrow$ $\widehat{A}$=B-C=180-50-55=75