An exterior angle of a triangle is $105^{\circ}$ and its two interior opposite angles are equal. Each of these equal angles is

  • A
    ${37 \frac{1}{2}}^{\circ}$
  • B
    $75^{\circ}$
  • C
    ${72 \frac{1}{2}}^{\circ}$
  • D
    ${52 \frac{1}{2}}^{\circ}$

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In the given figure,in $\Delta PQR$,$\angle Q > \angle R$. $PM \perp QR$ and $PA$ is the bisector of $\angle QPR$. Prove that $\angle APM = \frac{1}{2}(\angle Q - \angle R)$.

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