In the given figure, AM ⊥ BC and AN is the bisector of ∠A. Then ∠MAN is
In given figure,
∠BAC+∠ABC+∠ACB=180∘(Angles of same triangle)⇒∠BAC+65∘+33∘=180∘⇒∠BAC=82∘⇒∠CAN=∠BAC2=41∘(Given that AN is an angle bisector).Now, ∠CAN+∠ANC+∠ACN=180∘(Angles of same triangle)⇒41∘+∠ANC+33∘=180∘⇒∠ANC=106∘∠ANC+∠ANM=180∘(Linear pair)⇒106∘+∠ANM=180∘⇒∠ANM=74∘∠ANM+∠AMN+∠MAN=180∘⇒74∘+90∘+∠MAN=180∘⇒∠MAN=16∘.