a) Ta có:

$\widehat{A}$ + $\widehat{B}$ + $\widehat{C}$ = $180^o$ (định lý tổng 3 góc trong tam giác)

$35^o$ + $72^o$ + $\widehat{C}$ = $180^o$

$\widehat{C}$ = $180^o$ - ($35^o$ + $72^o$)

$\widehat{C}$ = $73^o$

Mặt khác, ta có:

$\widehat{A}$ < $\widehat{B}$ < $\widehat{C}$

$\Rightarrow$ BC < AC < AB

b) Gọi góc ngoài tại đỉnh A là $\widehat{A1}$

Ta có:

$\widehat{A}$ + $\widehat{A1}$ = $180^o$ (2 góc kề bù)

$\widehat{A}$ + $110^o$ = $180^o$

$\widehat{A}$ = $180^o$ - $110^o$

$\widehat{A}$ = $90^o$

Ta có:

$\widehat{A}$ + $\widehat{B}$ + $\widehat{C}$ = $180^o$ (định lý tổng 3 góc trong tam giác)

$90^o$ + $42^o$ + $\widehat{C}$ = $180^o$

$\widehat{C}$ = $180^o$ - ($90^o$ + $42^o$)

$\widehat{C}$ = $48^o$

Mặt khác, ta có:

$\widehat{B}$ < $\widehat{C}$ < $\widehat{A}$

$\Rightarrow$ AC < AB < BC

c) Vì $\triangle$ABC cân tại A nên $\widehat{B}$ = $\widehat{C}$

Ta có:

$\widehat{A}$ + $\widehat{B}$ + $\widehat{C}$ = $180^o$ (định lý tổng 3 góc trong tam giác)

$30^o$ + $\widehat{B}$ + $\widehat{C}$ = $180^o$

$\widehat{B}$ + $\widehat{C}$ = $180^o$ - $30^o$

$\widehat{B}$ + $\widehat{C}$ = $150^o$

$\Rightarrow$ $\widehat{B}$ = $\widehat{C}$ = $150^o$ : 2 = $75^o$

Mặt khác, ta có:

$\widehat{A}$ < $\widehat{B}$ = $\widehat{C}$

$\Rightarrow$ BC < AC = AB

d) Ta có:

$\widehat{A}$ + $\widehat{B}$ + $\widehat{C}$ = $180^o$ (định lý tổng 3 góc trong tam giác)

$50^o$ + $\widehat{B}$ + $\widehat{C}$ = $180^o$

$\widehat{B}$ + $\widehat{C}$ = $180^o$ - $50^o$

$\widehat{B}$ + $\widehat{C}$ = $130^o$

Vì số đo của $\widehat{B}$ và $\widehat{C}$ tỉ lệ với 6;7 nên:

$\dfrac{\widehat{B}}{6}$ = $\dfrac{\widehat{C}}{7}$

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

$\dfrac{\widehat{B}}{6}$ = $\dfrac{\widehat{C}}{7}$ = $\dfrac{\widehat{B}+\widehat{C}}{6+7}$ = $\dfrac{130^o}{13}$ = $10^o$

Suy ra $\widehat{B}$ = $10^o$ . 6 = $60^o$

$\widehat{C}$ = $10^o$ . 7 = $70^o$

Mặt khác, ta có:

$\widehat{A}$ < $\widehat{B}$ < $\widehat{C}$

$\Rightarrow$ BC < AC < AB