@iambis
Bài 5.
a)
$\longrightarrow$ Xét $\triangle$ ADB, ta có:
$\Rightarrow$ $\widehat{BAD}$ + $\widehat{B}$ + $\widehat{ADB}$ = $180^o$ (tổng 3 góc trong 1 $\triangle$)
$\Rightarrow$ $60^o$ + $50^o$ + $\widehat{ADB}$ = $180^o$
$\Rightarrow$ $\widehat{ADB}$ = $180^o$ - ( $60^o$ + $50^o$ )
$\Rightarrow$ $\widehat{ADB}$ = $70^o$
$\longrightarrow$ Ta có: $\widehat{ADB}$ + $\widehat{ADC}$ = $180^o$ (2 góc kề bù)
$\Rightarrow$ $70^o$ + $\widehat{ADC}$ = $180^o$
$\Rightarrow$ $\widehat{ADC}$ = $180^o$ - $70^o$
$\Rightarrow$ $\widehat{ADC}$ = $110^o$
b)
$\longrightarrow$ Xét $\triangle$ ADC, ta có:
$\Rightarrow$ $\widehat{ADC}$ + $\widehat{C}$ + $\widehat{DAC}$ = $180^o$ (tổng 3 góc trong 1 $\triangle$)
$\Rightarrow$ $110^o$ + $40^o$ + $\widehat{DAC}$ = $180^o$
$\Rightarrow$ $\widehat{DAC}$ = $180^o$ - ( $110^o$ + $40^o$ )
$\Rightarrow$ $\widehat{ADB}$ = $30^o$
~ Gửi e nha, cho a xin ctlhn nha ::>_<:: ~
