BITSAT 2019 Paper

(d) Let height of the tower be h m and distance between tower and cliff be x m.

∴ CD = h, BD = x
In ΔABD, tan 45° =

AB

BD

or 1 =

50

x

x = 50 .......(i)
In ΔAEC
tan 30° =

AE

EC

=

AB−EB

EC

=

AB−DC

BD

(∵EB = DC, EC = BD)

1

√3

=

50−h

x

or x = 50 √3 - h√3
or 50 = 50√3 - h√3 [From (i), x = 50]
or h√3 = 50√3 - 50
or h =

50(√3−1)

√3

= 50(1−

1

√3

)
∴ h = 50 (1−

√3

3

)