the angles of depression of the top and bottom of an 8m tall building from the top and bottom of - Maths - Some Applications of Trigonometry

Question

the angles of depression of the top and bottom of an 8m tall
building from the top and bottom of a multistoreyed buildin are 30
and 45 degrees respectively find the height of the multistoreyed
building and the distance between the twoo buildings please anser
me fast today is my test

Gursheen kaur. · Accepted Answer

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° (Corresponding angles)

∠FBC = ∠BCA = 45° (Corresponding angles)

Now,

In Δ ACB,

$\Rightarrow \tan 45^{0} = \frac{AB}{AC} [∵ \tan θ = \frac{Perpendicular}{Base}] \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h ...... (i)$

In Δ BDE,

$\Rightarrow \tan 30^{0} = \frac{BE}{ED} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h - 8}{x} \Rightarrow x = \sqrt{3} (h - 8) ............ (i i)$

From (i) and (ii), we get,

$\Rightarrow h = \sqrt{3} h - 8 \sqrt{3} \Rightarrow \sqrt{3} h - h = 8 \sqrt{3} \Rightarrow h (\sqrt{3} - 1) = 8 \sqrt{3} \Rightarrow h = \frac{8 \sqrt{3}}{\sqrt{3} - 1} \Rightarrow h = \frac{8 \sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{(3 - 1)} = \frac{8 \sqrt{3} (\sqrt{3} + 1)}{2} = 4 \sqrt{3} (\sqrt{3} + 1) \Rightarrow h = (12 + 4 \sqrt{3}) m$

Distance between the two building

$x = (12 + 4 \sqrt{3}) m [F r o m (i)]$

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the angles of depression of the top and bottom of an 8m tall building from the top and bottom of a multistoreyed buildin are 30 and 45 degrees respectively. find the height of the multistoreyed building and the distance between the twoo buildings. please anser me fast today is my test.