Ai raspunsul in atasament
(ABD)∩(ACD)=AD
fie M∈AD, AM≡MD
BN si CM⊥AD, m∡((ABD),(ACD))=m∡BM, MC
apoii am folosit aria exprimata prin 2 formule BC*MN/2= BM *MC *sin ∡(BMC)/2
de unde am aflat sin∡(BMC) apoi cu form fundam. a trifgonom. a, aflat cosinusul, cerinta
