Xét $ΔDEF$:
$\widehat{D}+\widehat{E}+\widehat{F}=180^\circ$ (tổng 3 góc trong 1 Δ)
$→\widehat{F}=180^\circ-\widehat{D}-\widehat{E}$
$→\widehat{F}=180^\circ-90^\circ-60^\circ=30^\circ$
Xét $ΔFMH$:
$\widehat{F}+\widehat{MHF}+\widehat{FMH}=180^\circ$ (tổng 3 góc trong 1 Δ)
$→\widehat{FMH}=180^\circ-\widehat{F}-\widehat{MHF}$
$→\widehat{FMH}=180^\circ-30^\circ-90^\circ=60^\circ$
Vậy $\widehat{FMH}=60^\circ$