Answer-
[tex]\boxed{\boxed{m\angle EBC=65^{\circ}15'}}\\\\\boxed{\boxed{AC=20\ in}}[/tex]
Solution-
The ΔABC is an isosceles triangle, as AB = BC
And m∠FEC=90°, so m∠BEC=90° (as they are complimentary angle)
Hence, BE is an altitude to the base.
From the properties of an isosceles triangle (where base is the side which is not equal to any other side), we know that
The altitude drawn to the base is the median and the angle bisector.
So,
[tex]\Rightarrow m\angle ABC=2\times m\angle EBC[/tex]
[tex]\Rightarrow m\angle EBC=\dfrac{1}{2}m\angle ABC=\dfrac{1}{2}\times 130^{\circ}30'[/tex]
[tex]\Rightarrow m\angle EBC=65^{\circ}15'[/tex]
And also,
[tex]AC=2\times AE=2\times 10=20\ in[/tex]
