from the lower level ground the angle of elevation is such that the tangents are 5÷12 and if moved 192 m towards the tower that tangents become 3÷5 . then find the height of the tower
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Let AB be the tower of height h metres. Let AD=x metres, CD=192 metres.
tanα = , tanβ =
In ∆BAC,
tanα = = ……………………. (i)
In ∆DAB,
tanβ = = or x= ……………………… (ii)
Using (ii) in (i)
=
5= 12h
2880 +20h =36h
16h = 2880 or h= 180
Hence, the height of the tower is 180 metres.
OR
Let AB and CD be two towers of height h m and 60 m respectively.
AC=140m and ÐBDE =300.
In ∆DEB,
tan 30° =
BE == 80.83m
Thus, the height of the first tower is
AB= AE+BE = CD+BE =60+80.83 = 140.83m
Explanation:
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