The location of points E and F has not been specified. However if it is presumed that D lies on AC and DE is drawn perpendicular to AC and F lies on BC such that angle DEF is a right angle , the figure drawn to show all these points will be as shown below.

Since angle DEF is a right angle, EF will be parallel to AC and therefore in Triangle BEF, BF= BE. Consequently AE = FC

Consider Right Triangle ADE, in which angle A is 60 degrees and side AD is 5cms. Hence AE= 5Sec60 = 10 cms and DE= 5tan 60= 5√3 cms

Now draw FG perpendicular to AC. DEEG will be a rectangle

In Right Triangle GCF since FC= 5cms and angle C is 60 degrees, GC=10 cos60 = 5cms.

This makes DG = 8 cms = EF

Then using Pythagoras theorem DF= √139, as shown in the figure above

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