In Fig. 9.37, AM ⊥ BC and AN is the bisector of ∠A. If ∠B =65° and ∠C =33°, find ∠MAN.

Question

In Fig. 9.37, AM &#x22A5; BC and AN is the bisector of &#x2220; A. If &#x2220; B =65&#xB0; and &#x2220; C =33&#xB0;, find &#x2220; MAN.

Philoid · Accepted Answer

Given,

AM perpendicular to BC

AN is bisector of ∠A

Therefore, ∠NAC = ∠NAB

In

∠A + ∠B + ∠C = 180^o

∠A + 65^o + 33^o = 180^o

∠A = 180^o – 98^o

= 82^o

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

In

∠AMB + ∠MAB + ∠ABM = 180^o

90^o + ∠MAB + 65^o = 180^o

∠MAB + 155^o = 180⁰

∠MAB = 25^o

Therefore,

∠MAB + ∠MAN = ∠BAN

25^o + ∠MAN = 41^o

∠MAN = 41^o – 25^o

= 16^o