Xét $ΔABC(\widehat{A}=90^o):$
$sin_C=\dfrac{AB}{BC}$
$⇒AB=BC.sin{C}=4.sin_{15^o}=\sqrt{6}-\sqrt{2}(cm)$
Ta có $AM$ trung tuyến
$⇒BM=\dfrac{1}{2}BC=\dfrac{1}{2}.4=2(cm)$
$\widehat{C}+\widehat{B}=90^o$
$⇒\widehat{B}=90^o-\widehat{C}=90^o-15^o=75^o$
Xét $ΔAHB(\widehat{AHB}=90^o):$
$⇒cos{B}=\dfrac{BH}{AB}=$
$⇒BH=AB.cos{B}=(\sqrt{6}-\sqrt{2}).cos{75}=2-\sqrt{3}(cm)$
$BH^2+AH^2=AB^2$ $(Ptg)$
$⇒AH=\sqrt{AB^2-BH^2}=\sqrt{(\sqrt{6}-\sqrt{2})^2-(2-\sqrt{3})^2}=1(cm)$
$BH+HM=BM$
$⇒HM=BM-BH=2-(2-\sqrt{3})=\sqrt{3}(cm)$
Xét $ΔAHM(\widehat{AHM}=90^o):$
$AM^2=AH^2+HM^2$
$⇒AM=\sqrt{AH^2+HM^2}=\sqrt{1^2+(\sqrt{3})^2}=2(cm)$
$⇒sin_{AMH}=\dfrac{AH}{AM}=\dfrac{1}{2}$
$⇒\widehat{AMH}=30^o$
