# An observer

Let height of the tower CD be h m.

Consider ΔDCB,

tan45° = DC/CB

1 = h/CB

CB = h

Consider ΔDCA,

tan30° = DC/CA

1/√3 = h/CA

Since CA = CB + AB = h + 20

1/√3 = h/h + 20

h + 20 = √3h

√3h – h = 20

h (√3 – 1) = 20

h = 20/ (√3 – 1)

h = 10 + 10√3

h = 27.32 m

Ans. The height of the tower is 27.32 m.

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