In the given figure, AM ⊥ BC and AN is the bisector of ∠A. Then ∠MAN is

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In the given figure, AM $⊥$ BC and AN is the bisector of $∠$ A. Then $∠$ MAN is

Options:

A .

32 {\frac{1}{2}}^{\circ}

B .

16 {\frac{1}{2}}^{\circ}

C .

16^{\circ}

D .

32^{\circ}

Answer: Option C
:
C

In given figure,
$∠ B A C + ∠ A B C + ∠ A C B = 180^{\circ} (Angles of same triangle) \Rightarrow ∠ B A C + 65^{\circ} + 33^{\circ} = 180^{\circ} \Rightarrow ∠ B A C = 82^{\circ}$
$∠ C A N = \frac{∠ B A C}{2} = 41^{\circ} (Given that AN is an angle bisector) . Now, ∠ C A N + ∠ A N C + ∠ A C N = 180^{\circ} (Angles of same triangle) \Rightarrow 41^{\circ} + ∠ A N C + 33^{\circ} = 180^{\circ} \Rightarrow ∠ A N C = 106^{\circ}$
$∠ A N C + ∠ A N M = 180^{\circ} (Linear pair) \Rightarrow 106^{\circ} + ∠ A N M = 180^{\circ} \Rightarrow ∠ A N M = 74^{\circ}$
$∠ A N M + ∠ A M N + ∠ M A N = 180^{\circ} \Rightarrow 74^{\circ} + 90^{\circ} + ∠ M A N = 180^{\circ} \Rightarrow ∠ M A N = 16^{\circ} .$

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