In Fig., AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and

1.	In Fig., AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.
Answer» Given, AM perpendicular to BC AN is bisector of ∠A Therefore, ∠NAC = ∠NAB In ΔABC ∠A + ∠B + ∠C = 180° ∠A + 65° + 33° = 180° ∠A = 180° – 98° = 82° ∠NAC = ∠NAB = 41°(Therefore, AN is bisector of ∠A) In ΔAMB ∠AMB + ∠MAB + ∠ABM = 180° 90° + ∠MAB + 65° = 180° ∠MAB + 155° = 180° ∠MAB = 25° Therefore, ∠MAB + ∠MAN = ∠BAN 25° + ∠MAN = 41° ∠MAN = 41° – 25° = 16°

In Fig., AM ⊥ BC and AN is the bisector of ∠A. If ∠B = 65° and ∠C = 33°, find ∠MAN.

Answer»

Given,

AM perpendicular to BC

AN is bisector of ∠A

Therefore, ∠NAC = ∠NAB

In ΔABC

∠A + ∠B + ∠C = 180°

∠A + 65° + 33° = 180°

∠A = 180° – 98°

= 82°

∠NAC = ∠NAB = 41°(Therefore, AN is bisector of ∠A)

In ΔAMB

∠AMB + ∠MAB + ∠ABM = 180°

90° + ∠MAB + 65° = 180°

∠MAB + 155° = 180°

∠MAB = 25°

Therefore,

∠MAB + ∠MAN = ∠BAN

25° + ∠MAN = 41°

∠MAN = 41° – 25°

= 16°