The angles of dep

Let’s take PQ = 60√3 be the cliff and

CD be the tower

In right ∆PQR,

Tan 60° = PQ/QR

√3 = 60√3/QR

√3 QR = 60√3

QR = 60√3/√3 = 60 m

In right ∆PTS

Tan 45° = PT/ST

1 = PT/60

PT = 60 m

Therefore;

Height of the tower,

RS = QT

= PQ – PT

= 60√3 – 60 = 60(√3 – 1)

= 60(1.73 – 1)

= 60(0.73)

= 43.8 m