Question

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

Answer 1

Let AD be the multistoried building of height hm. And the angle of depression of the top and bottom are 30° and 45°. We assume that BE = 8, CD = 8 and BC = x, ED = x and AC = h − 8. Here we have to find height and distance of the building.

We use trigonometric ratio.

In ΔAED,

Hence the required height is 4(3 + √3) meter and distance is 4(3 + √3) meter.

Answer 2

Let AB and CD be the multi-storeyed building and the building respectively. 

Let the height of the multi-storeyed building = h m and 

The distance between the two buildings = x m. 

AE = CD = 8 m [Given] 

BE = AB – AE = (h – 8) m 

And AC = DE = x m [Given] 

Also, ∠FBD = ∠BDE = 30° (Alternate angles) 

∠FBC = ∠BCA = 45° (Alternate angles) 

Now, 

In Δ ACB,

tan 45° = \(\frac{AB}{AC}[∵tanθ=\frac{perpendicular}{Base}]\)

\(1=\frac{h}{x}\)

x = h ... (i)

In Δ BDE,

tan 30° = \(\frac{BE}{ED}\)

\(\frac{1}{\sqrt{3}}= \frac{h-8}{x}\)

x = \(\sqrt{3}(h - 8)\) ... (ii) 

From (i) and (ii), 

We get, h = √3h - 8√3 

√3h – h = 8√3 

h(√3 - 1) = 8√3 

h- = 4√3(√3 + 1) 

h = 12 + 4√3 m 

Distance between the two building

\(x = (12+4\sqrt{3})m\) [From (i)]