Distance between two poles is BD=15m
Hence, CE=15m
Height of second pole, AB=24m
Let, height of first pole be CD=hm
Hence, BE=hm
Therefore, AE=24−hm
In right angled triangle AEC,
tan30⁰=AE/CE
→1/√3=(24−h)/15
→15=24√3−√3h
→√3h=24√3−15
→h=(24−15/√3)=(24−5√3)=24−5×1.732=24−8.66=15.34m
Therefore, h=15.34m