Xét tam giác ABC vuông tại A có:
AC = AB.tan\(\widehat{ABC}\) = 4,5467. tan59o12'37'' \(\approx\) 7,6303 (cm)
BC = \(\dfrac{AB}{cos\widehat{ABC}}=\dfrac{4,5467}{cos59^o12'37"}\)\(\approx\) 8,8822 (cm)
Do AM là trung tuyến tam giác ABC
=> AM = MC = MB = \(\dfrac{1}{2}\)AB \(\approx\) \(\dfrac{1}{2}\).8,8822 = 4,4411 (cm)
Kẻ NK ⊥ AC => NK // AB (cùng vuông góc AC)
AN là phân giác trong tam giác ABC =>\(\widehat{NAB}=\widehat{NAC}=45^o\); \(\dfrac{NC}{NB}=\dfrac{AC}{AB}=\dfrac{7,6303}{4,5467}\) => \(\dfrac{NC}{BC}=\dfrac{7,6303}{12,177}\) (*)
Do NK // AB (cmt) => \(\dfrac{NK}{AB}=\) \(\dfrac{NC}{BC}=\dfrac{7,6303}{12,177}\)
=> NK = \(\dfrac{7,6303}{12,177}.AB=\dfrac{7,6303}{12,177}.4,5467\approx2,849\) (cm)
Xét tam giác ANK vuông tại K có: AN = \(\dfrac{NK}{sin\widehat{NAK}}=\dfrac{2,849}{sin45^o}\approx2,015\left(cm\right)\)
Kẻ AH ⊥ BC. Xét tam giác ABC vuông tại A có: AH.BC = AB.AC
=> AH = \(\dfrac{AB.AC}{BC}=\dfrac{4,5467.7,6303}{8,8822}\approx3,9509\left(cm\right)\)
Từ (*) => NC = \(\dfrac{7,6303}{12,177}.BC=\dfrac{7,6303}{12,177}.8,8822\approx5,5657\) (cm)
=> MN = NC - MC = 5,5657 - 4,4411 = 1,1246 (cm)
=> SAMN = \(\dfrac{1}{2}\).AH.MN = \(\dfrac{1}{2}\).3,9509.1,1246 \(\approx\) 2,2216 (cm2)
