The angles of depression of the top and the bottom of a 8 m tall building from the top of a multi-storeyed building are 30° and 45° respectively. Find the height of the multi storeyed building and the distance between the two buildings, correct to two decimal places.

Let the height of multi storeyed building (AB) be h meters and the distance between two buildings be x meters.

From figure,

∠ADB = ∠XAD = 45° (Alternate angles are equal)
∠AFE = ∠XAF = 30° (Alternate angles are equal)

As opposite sides of a rectangle are equal

FE = DB = x
BE = DF = 8

AE = AB - BE = h - 8

Considering right angled △ADB, we get

$\Rightarrow \text{tan 45°} = \dfrac{AB}{DB} \\[1em] \Rightarrow 1 = \dfrac{h}{x} \\[1em] \Rightarrow x = h \text{ ……(Eq 1)}$

Considering right angled △AFE, we get

$\Rightarrow \text{tan 30°} = \dfrac{AE}{FE} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h - 8}{x} \\[1em] \Rightarrow x = \sqrt{3}(h - 8)$

Putting value of x from Eq 1 in above equation,

$\Rightarrow h = \sqrt{3}h - 8\sqrt{3} \\[1em] \Rightarrow \sqrt{3}h - h = 8 \sqrt{3} \\[1em] \Rightarrow 0.732\text{ h} = 13.856 \\[1em] \Rightarrow h = \dfrac{13.856}{0.732} \\[1em] \Rightarrow h = 18.93$

x = h = 18.93

Hence, the height of multi storeyed building is 18.93 meters and the distance between the two buildings is 18.93 meters.