Solution:

Consider AB as the height and CD as the building

The angles of depression from A to C and D are 30⁰ and 45⁰

∠ACE = 30⁰ and ∠ADB = 45⁰

CD = 8 m

Take AB = h and BD = x

From the point C

Construct CE parallel to DB

CE = DB = x

EB = CD = 8 m

AR = AB – EB = h – 8

In right triangle ADB

tan θ = AB/DB

Substituting the values

tan 45⁰ = h/x

So we get

1 = h/x

x = h

In right triangle ACE

tan 30⁰ = AE/CE

Substituting the values

1/√3 = (h – 8)/ h

By further calculation

h = √3h – 8√3

So we get

√3h – h = 8√3

h (√3 – 1) = 8√3

h = 8√3/(√3 – 1)

Multiply and divide by √3 + 1

h = 8√3/ (√3 – 1) × (√3 + 1)/ (√3 + 1)

h = 8 (3 + √3)/ (3 – 1)

Here

h = 8 (3 + 1.732)/ 2

h = 4 × 4.732

h = 18.928

h = 18.93 m

x = h = 18.93 m

Here

Height of multi-storeyed building = 18.93 m

Distance between the two buildings = 18.93 m