Solution:

Let the height of the tower(AB) = h m

Let the length of the shorter shadow be x m

Then, the longer shadow is (10 + x)m

So, from fig. In ΔABC

tan 60^o = AB/BC

√3 = h/x

x = h/√3…. (i)

Next, in ΔABD

tan 45^o = AB/BD

1 = h/(10 + x)

10 + x = h

10 + (h/√3) = h [using (i)]

10√3 + h = √3h

h(√3 -1) =10√3

h = 10√3/ (√3 -1)

After rationalising the denominator, we have

h = [10√3 x (√3 + 1)]/ (3 – 1)

h = 5√3(√3 + 1)

h = 5(3 + √3) = 23.66 [√3 = 1.732]

Therefore, the height of the tower is 23.66 m.