The angel of the elevation of the top of a tower from the point a (on the ground) is 30°.on walking 50 metres towards the angel of the elevation is found to be 60°.Calculate: the height of the tower
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Step-by-step explanation:
In ΔADC
tan30
o
=
AC
DC
3
1
=
d
h
........(1)
d=
3
h
→d=
3
×10
3
=30m
In ΔBCD
tan60
o
=
BC
DC
3
=
d−20
h
.......(2)
on solving equation (1) & (2)
3
=
3
h−20
h
⇒3h−20
3
=h
2h=20
3
h=10
3
m
height of tower.
solution
Answered By
toppr
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