Sum of interior angles of a regular polygon is given as \((\mathrm{n}-2) 180\) for a pentagon, \(\mathrm{n}=5\) \((\mathrm{n}-2) 180=(5-2) \times 180=3 \times 180=\) \(540^{\circ}\)
if four of its angles are equal, then 4 (one angle) \(+60^{\circ}=540\)
\(\therefore\) size of one angle \(=\frac{540-60}{4}=120^{\circ}\)