Question

In triangle \( ABC , AM \perp BC \). AN the bisector of angle \( BAC \). If \( BC \) act as 1 a plane mirror and this triangle \( ABC \) is reflected along \( BC \), the figure so formed is given below. \( m \angle A B C=70^{\circ} \) and \( m \angle A C B=40^{\circ} \), Then, the value of \( \angle M D N \) is (in degree) A. 30 B. 45 C. 15 D. 55

Answer 1

M < ABC = 70°

\(\therefore\) \(\angle CBD\) = 90° - 70° = 20°

\(\therefore\) M < ACB = 4° C

\(\therefore\) \(\angle BCD\) = 90°- 40° = 50°

In \(\triangle\)BCD,

\(\angle BDC\) = 180° - \(\angle BCD\) - \(\angle CBD\)

= 180° -20° -50°

= 110°

\(\because\) AN is bisector of \(\angle A\)

\(\therefore\)  DN is bisector of \(\angle D\)

\(\therefore\) \(\angle BDN\) = \(\angle CDN\) = \(\cfrac{\angle D}{2}\) = \(\cfrac{110^o }{2} = 55^o\)

In \(\triangle\) NCD,

\(\angle CND\) = 180° - \(\angle NCD\) - \(\angle CDN\)

= 180° - 50° - 55°

= 180° - 105°

= 75°

In \(\triangle\) NMD,

\(\angle MDN\) = 180° - \(\angle NMD\) - \(\angle CND\)

= 180° - 90° - 75°

= 90°- 75°

= 15°

\(\therefore\) \(\angle MDN\) = 15°

Option (C) is correct.

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