In $\triangle ABC, BD \perp AC$ at D. E is a point on BC such that $\angle BEA = x^\circ$. If $\angle EAC = 62^\circ$ and $\angle EBD = 60^\circ$, then the value of $x$ is:

$92^\circ$

In \(\Delta \)AOD,

\(\angle\)AOD + \(\angle\)ADO + \(\angle\)DAO = 180

= \(\angle\)AOD + 90 + 62 = 180

= \(\angle\)AOD = 180 - 152

= \(\angle\)AOD = 28

Therefore, \(\angle\)AOD and \(\angle\)EOB are vertical opposite angles,

\(\angle\)AOD = \(\angle\)EOB = 28

In \(\Delta \)BOE,

= \(\angle\)BEO + \(\angle\)EOB + \(\angle\)OBE = 180

= x + 28 + 60 = 180

= x = 180 - 88

= x = 92

Therefore, x is \({92}^\circ\).