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Question:

The horizontal distance between two poles is 15m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30o. If the height of the second pole is 24m, find the height of the first pole. [Use √3=1.732]

Solution:

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