Correct Answer - C
`angle ADE =angle ADC +angle CDE =90^@+60^@ (because "Angles in a square and equilateral triangle")=150^@`
In `Delta ADE, AD=DE`
`therefore angle DAE =angle DEA =(1)/(2)(180^@-150^@)`
Similarly, `angle CEB =15^@`
`therefore angle AEB=60^@-(angle DEA +angle CEB)`
`=60^@-(15^@+15^@)=30^@`.