Đáp án:
Giải thích các bước giải:
a, $\triangle$ ABC có $\widehat{BAC}$ + $\widehat{ABC}$ + $\widehat{ACB}$ = $180^o$
=> $90^o$ + $60^o$ + $\widehat{ACB}$ = $180^o$
=> $\widehat{ACB}$ = $30^o$
b, Vì AD là tia phân giác $\widehat{CAB}$ => $\widehat{DAB}$ = $\widehat{CAB}$ : 2
=> $\widehat{DAB}$ = $90^o$ : 2 = $45^o$
- $\triangle$ ADB có: $\widehat{DAB}$ + $\widehat{DBA}$ + $\widehat{ADB}$ = $180^o$
=> $45^o$ + $60^o$ + $\widehat{ADB}$ = $180^o$
=> $\widehat{ADB}$ = $75^o$ (hay $\widehat{ADH}$ = $75^o$
c, Ta có: $\widehat{HAC}$ + $\widehat{ACB}$ = $90^o$ ($\triangle$ AHC ⊥ tại C)
$\widehat{ABC}$ + $\widehat{ACB}$ = $90^o$ ($\triangle$ ABC ⊥ tại A)
=> $\widehat{HAC}$ = $\widehat{ABC}$
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