Chủ đề này có 1 trả lời

[kienvuhoang]

Cho$\Delta ABC và điểm M$ thỏa mãn $\angle MAB=10;\angle MBA=20;\angle MCA=30;\angle MAC=40$

Tính $\angle MBC$

[khgisongsong]

tính được $\widehat{AMB}=150^{\circ},\widehat{AMC}=110^{\circ},\widehat{CMB}=100^{\circ},\widehat{BAC}=50^{\circ}$

đặt a=AM

tính được $MC=\frac{sin40^{\circ}}{sin30^{\circ}}.a$

$ AB=\frac{sin150^{\circ}}{sin20^{\circ}}.a$

$AC=\frac{sin110^{\circ}}{sin30^{\circ}}.a$

mà $BC=\sqrt{AB^2+AC^2-2AB.AC.cos50^{\circ}}$

suy ra $BC=\frac{a}{sin20^{\circ}.sin30^{\circ}}.\sqrt{sin^2150^{\circ}.sin^230^{\circ}+sin^2110^{\circ}.sin^220^{\circ}-2sin150^{\circ}.sin110^{\circ}.cos50^{\circ}.sin20^{\circ}.sin30^{\circ}}$

mà $sin^2150^{\circ}.sin^230^{\circ}+sin^2110^{\circ}.sin^220^{\circ}-2sin150^{\circ}.sin110^{\circ}.cos50^{\circ}.sin20^{\circ}.sin30^{\circ}$

=$sin^4(30^{\circ})+cos^2(-20^{\circ}).sin^220^{\circ}-2.sin^230^{\circ}.cos20^{\circ}.sin20^{\circ}.sin40^{\circ}$ (do $sin150^{\circ}=sin30^{\circ}$ và $cos50^{\circ}=sin40^{\circ}$)

=$sin^4(30^{\circ})+\frac{sin^240}{4}-sin^230.sin^240=\frac{1}{16}$ do $sin30=1/2$

suy ra  $BC=\frac{a}{sin20^{\circ}.sin30^{\circ}}.\frac{1}{4}$

có $ \frac{BC}{sin\widehat{BMC}}=\frac{MC}{sin\widehat{MBC}}$

suy ta $sin\widehat{MBC}=\frac{MC.sin100^{\circ}}{BC}=4.sin40^{\circ}.sin20^{\circ}.sin100^{\circ}=2(cos20^{\circ}-cos60^{\circ}).sin80^{\circ} =2cos20^{\circ}.sin80^{\circ}-2cos60^{\circ}.sin80^{\circ}=sin100^{\circ}+sin60^{\circ}-sin140^{\circ}-sin20^{\circ}$

$=sin60^{\circ}+(sin100^{\circ}-sin20^{\circ})-sin40^{\circ}$(do $sin140^{\circ}=sin40^{\circ}$)

$=sin60^{\circ}+2.cos60^{\circ}.sin40^{\circ}-sin40^{\circ} =sin60^{\circ}$ (do $cos60^{\circ}=\frac{1}{2}$)

$=>\widehat{MBC}=60^{\circ}$