Đáp án + giải thích các bước giải:

a) `sin \hat{BCA} =(BA)/(BC)`

`->BC=(BA)/(sin \hat{BCA})=20/(sin 15^0)≈77,28`

`tan \hat{BCA}=(AB)/(AC)`

`->AC=(AB)/(tan \hat{BCA})=20/(tan 15^0)≈74,64`

b) `sin \hat{BAC}=(BC)/(AC)`

`->BC=sin \hat{BAC}.AC=sin 60^0 . 5=(5\sqrt{3})/2`

`cos \hat{BAC}=(AB)/(AC)`

`->AB=cos \hat{BAC}.AC=cos 60^0 . 5 = 5/2 `

c) `sin \hat{A}=(AC)/(AB)`

`->AB=(AC)/(sin \hat{A})=12/(sin30^0)=24`

`cos \hat{A}=(BC)/(BA)`

`->BC=cos \hat{A} . BA=cos30^0 . 24=12\sqrt{3}`

d) `sin \hat{NPM}=(NM)/(NP)`

`->NM=sin\hat{NPM}.NP=sin37^0 . 14≈8,43`

`cos \hat{NPM}=(PM)/(PN)`

`->PM=cos \hat{NPM}.PN=cos 37^0 . 14≈11,18`

e) `sin \hat{PMN}=(NP)/(NM)`

`->NM=(NP)/(sin \hat{PMN})=20/(sin50^0)=26,11`

`tan \hat{PMN}=(PN)/(PM)`

`->PM=(PN)/(tan \hat{PMN})=20/(tan 50)=16,78`

f) `tan \hat{NMP}=(NP)/(NM)`

`->NP=tan \hat{NMP}.NM=tan 34^0 . 25=16,86`

`cos \hat{NMP}=(MN)/(MP)`

`->MP=(MN)/(cos \hat{NMP})=25/(cos 34^0)=30,16`