If one of the angles of a triangle is 74°, then the angle between the bisectors of the other two interior angles is :

127°

In \(\Delta \)ABC, such that \(\angle\)ACB = \({74}^\circ\) and bisectors \(\angle\)A and \(\angle\)B meet at D and E.

Now, In \(\Delta \)ABC,

\(\angle\)ACB + \(\angle\)ABC + \(\angle\)BAC = \({180}^\circ\)

= \({74}^\circ\) + \(\angle\)ABC + \(\angle\)BAC = \({180}^\circ\)

= \(\angle\)ABC + \(\angle\)BAC = \({180}^\circ\) - \({74}^\circ\) = \({106}^\circ\)

= \(\frac{1}{2}\)(\(\angle\)ABC + \(\angle\)BAC) = \({53}^\circ\)

= \(\angle\)PBA + \(\angle\)BAP = \({53}^\circ\)

Now, In \(\Delta \)PAB,

\(\angle\)PBA + \(\angle\)BAP + \(\angle\)BPA = \({180}^\circ\)

= \({53}^\circ\) + \(\angle\)BPA = \({180}^\circ\)

= \(\angle\)BPA = \({180}^\circ\) - \({53}^\circ\) = \({127}^\circ\)

Therefore, \(\angle\)BPA = \({127}^\circ\).